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Show the following: $\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+k)}=\dfrac{H_k}k$.
For each $n\in\Bbb N$ with $n\geq1$ is $\displaystyle H_n:=\sum_{k=1} ^n\dfrac 1k$ the $n$*-th partial sum of the harmonic series.* $k\in\Bbb N$ with $k\geq1$. Show that $\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+k)}=\dfrac{H_k}k$
How would I go about soving this? I've been doing math all day long and my head is about to explode. But I need this for tomorrow. Help please.
|
Note that $$\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right).$$
|
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How is it possible calculate Volumes from Adding together Areas? You can't find the volume for a $2D$ object, but by finding the area of say a square $A$ and multiply by any height $H$ you get the volume of cube with base $A$ and height $H$.
Here's my question. Is the volume of an object made of a very large amount of $2D$ surfaces stacked over each other? If yes, how is that possible? Since A $2D$ surface has thickness of $0$ isn't it like adding an infinite amount of zeroes and expecting to get a number that isn't zero?
|
I think this is just a matter of conceptualisation. But Calculus is better because it gives us the concept and the language to do the calculations.
Another way is instead of thinking that a 2D area has zero height, I would like to think that height is undefined in the 3rd dimension. Then when you multiply by the height, you give the physical object a 3rd dimension and the new definition is of a volume.
We need to consider the physical dimension or unit of measure that you are multiplying or "adding". For example,
*
*same dimension e.g. 1 unicorn + 1 unicorn = 2 unicorns;
*certain pre-requisites must be stated outside of the mathematical statement e.g. you cannot simply add density together normally without specifying you how do it physically;
*adding generates a new dimension, as in area into volume.
So in the case of area multiply by height, you are not really stacking sheets of zero height paper, but rather "extruding" the paper.
|
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Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$ Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself)
I could find a geometric solution in Wikipedia, but is there any solution that doesnt require drawing something?
Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change.
Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?
|
The differential equation
$$f''=-f$$
has a unique solution for given initial conditions $f(0)=x_0$, $f'(0)=y_0$. To show uniqueness assume that $g$ is another solution for the same initial values. Then $h:=f-g$ also satisfies $h''=-h$, and $h(0)=h'(0)=0$. But then
$$\frac\partial{\partial t}(h(t)^2+h'(t)^2)=2h(t)h'(t)+2h'(t)h''(t)=2h(t)h'(t)-2h'(t)h(t)=0,
$$
and it follows that $h(t)=0$ must hold for all $x$, so $f=g$.
This unique solution is
$$f(t)=y_0\sin t+x_0\cos t,
$$
as one checks immediately.
Now we can apply this to the function $f(t)=\sin(t+\beta)$. It satisfies
$f''(t)=-f(t)$, $f(0)=\sin\beta$, $f'(0)=\cos\beta$, hence we must have
$$f(t)=\cos\beta\sin t+\sin\beta\cos t
.$$
This may seem to come out of the blue at first, but if one sets $x(t)=f(t)$, $y(t)=f'(t)$, then $f''=-f$ becomes
$$\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}y\\-x\end{pmatrix},
$$
which describes a motion on a circle around the origin (the velocity is perpendicular to the vector from the origin to the current point), and of course $\sin$ and $\cos$ are all about circles. The calculation done above for $h$ shows that indeed $x(t)^2+y(t)^2$ is constant.
|
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Prove the limit $\mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin (x)}} = 1$ Prove the following limit:
$$\mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin (x)}} = 1$$
I can use limits arithmatic, squeezing principle, "well-known" limits etc..
We didn't learn Lopital law so I can't use it.
I tried to use the above tools, but didn't reach anything significant.
I'm suspecting it's involving trigonometric/logarithmic magic I'm not able to see.
|
EDIT: Note that on $(0,1)$, $x^a < x^b$ whenever $a>b$. The adjustment below reflects this. I give a shout out to Baranovskiy for point out this error.
I have an alternative proof. Note that $1 = x^0 \geq x^{\sin(x)} \geq x^x$ in some neighbourhood $(0,\delta)$. What does this tell us? Then squeeze.
|
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Real life examples of commutative but non-associative operations I've been trying to find ways to explain to people why associativity is important.
Subtraction is a good example of something that isn't associative, but it is not commutative.
So the best I could come up with is paper-rock-scissors; the operation takes two inputs and puts out the winner (assuming they are different).
So (paper rock) scissors= paper scissors = scissors,
But paper (rock scissors)= paper rock = paper.
This is a good example because it shows that associativity matters even outside of math.
What other real-life examples are there of commutative but non-associative operations? Preferably those with as little necessary math background as possible.
|
The averaging operation, defined by $$a\oplus b= \frac{a+b}2$$ is commutative but not associative.
|
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|
Simple Proof question Image : http://postimg.org/image/dkn0d5uen/
I'm studying Spivak's calculus and I have a really simple question :
I'm only in the first chapter on "The basic properties of numbers"
So far, we have the following propostion
P1 : (a+b)+c=a+(b+c)
P2 : a+0=0+a=a
P3 : a+(-a)=(-a)+a=0
Now, he tries to prove P2 (He doesn't do it for P3, so it's granted)
He also says :
"The proof of this assertion involves nothing more than subtracting a from both sides of the equation, in other word, adding -a to both sides." Now, that I understand
"as the following detailled proof shows, all three properties P1-P3 must be used to justify this operation." That I don't understand. First, how can you use a proof of something you haven't proven ? Second, when he says all three properties to justify this operation, he means to substract "a" from both sides, right ? If so, I don't understand how they (properties) can be used ...
He starts with this :
If a+x=a
then (-a)+(a+x)=(-a)+a=0
hence ((-a)+a)+x=0
hence 0+x=0
hence x=0
My comments : For the first line, he starts with the assertion that an equation a+x=a exists. Now, he substract "a" from borth sides and with property 3 the right hand sides equals 0. With property 1 we regroup and cancel with property 3.Now we have 0+x=0 and we subtract zero from both sides to have x=0. Where is property 2 used ? How is subtracting "a" from both sides proven with all three properties ?
Thank you
|
Spivak wants to show that zero is the unique additive identity on $\mathbb{R}$. That is, he want to prove that if we have $a+x=a$ then $x$ must identical to zero. He assumes P1, P2 and P3 to prove this. In particular, he uses P2 in the last step. If $0+x=0$ then using P2 we can conclude that $x=0$ without P2 we can not conclude this.
|
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RSA cryptosystem with special prime
Let $p < 2^{1000}$ and $q=3 \cdot 2^n - 1$ for $500 < n < 1000$ be
primes and set $n=pq$ to be the modulus of the RSA cryptosystem. Find
an attack on this system and how many operations that are required to
succeed.
My attempt at a solution: Set $m=pq$ and compute $d_n = \gcd(m, 3 \cdot 2^n - 1)$ for $500 < n <1000$ until we find a value $n=k$ such that $d_k>1$, then $q=(3 \cdot 2^k - 1)|m$ and we have cracked the system.
Is this correct? I saw a solution sketch for this using the same method iterating for $1\leq n \leq 1000$. Is it really necessary to test the first 500 values from the way this question is stated?
|
Your approach is correct; if the set of possible values of $q$ is so much restricted, the easiest solution is to search it exhaustively.
The tricky solution is to check the candidates for $q$ for primality first (this can be done once, even if we're trying to break multiple instances of the cryptosystem with different $p$ and $q$) to discover that the only case when it is prime occurs for $n=827$. Thus, the tricky guy could just plainly say: "$q=3\cdot 2^{827}-1$ and $p=n/q$", without doing any computation at all :-)
|
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Help on basic set theory question. Prove or Disprove: For every two sets $A$ and $B$, $(A\cup B)-B=A$.
I believed it was true, so first I showed that $(A\cup B)-B$ is a subset of $A$. My question is how do I prove that $A$ is a subset of $(A\cup B)-B$?
What I have first is what follows:
Suppose there exists an arbitrary element $x$ in $A$.
If $x$ is in $A$, then $x$ is not in $B$
From here, I'm stuck.
|
HINT: You are implicitly assuming that $A\cap B=\varnothing$.
|
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How does $\sin(x-2\pi) = \sin(x)$? How does $\sin(x-2\pi) = \sin(x)$? Is it so that you can split $\sin(x-2\pi)$ into $\sin(x) - \sin(2\pi)$ and that equals $\sin(x) - 0 = \sin(x)$? Please help. Thank you
|
Sine is a periodic function with period 2$\pi$. Perhaps the easiest way to see this is to use the formula $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$
|
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Volume of Solid Bounded by the cylinders $y=x^2$ and $y=z^2$ and the plane $y=1$ Find the volume of the solid bounded by the cylinders $y=x^2, y=z^2$ and the plane $y=1$
I think the integral should be: $$\int_0^1\int_{-\sqrt y}^\sqrt y\int_{-\sqrt y}^\sqrt y\ dx\,dz\,dy$$
Could someone tell me if this is correct?
|
This is not quite right. The region of integration there is not curved but straight like this:
Instead, maybe to
use cylindrical coordinates.
Also please look at this question here: Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer).
|
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work rate problem. GMAT -related I'm looking for a good formula/system to use for these problems. Too often I'm just relying on raw intuition and it takes me too much time to solve these questions. Is there a good starting place to solve these problems? What's like a good step 1 and step 2?
Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
So I don't know a good way to start. I thought output = rate * time but what is it here?
Let me think....
So rate * time = output. Okay. What else? Well, rates add onto each other without synergy... this is an assumption about rate problems on the GMAT. So if Machine A completes a 1/12 of a job in 1 hour and machine B completes 2/12 of a job in 1 hour, the two machines combined complete 3/12 of a job in 1 hour.
Is it 6 * r * 12 = 1
72r = 1
r = 1/72??
Why does that makes sense? Can I just arbitrarily make the job = 1?
So 1/72 is the rate of 1 machine.
x * 1/72 * 8 = 1
x = 9
So the answer is 9-6 = 3
What's a good way to think about this?
Someone, in a not very helpful way, suggested multiplying 6 * 12... but that doesn't explain to me what to do.
The answer is 3.
|
Any time you have multiple machines working together, you add their rate to get the total rate. In this case, the machines all have the same rate, so their total rate for the 6 machines is just 6r. We know that rate*time=work, so we know that 6r*12days=W (W can represent the job they complete)
Now we want to know how many more machines need to be added to complete the job in 8 days. We can represent that like this xr*8days=W. I find it easier to start by just making it x then subtracting the 6 later to avoid additional needless computation.
since, they are both completing the same job, we know that xr*8=6r*12 and solve for x
x=9, once we subtract the original 6 we get our answer 3
|
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How to calculate the interest amount per day I need to implement this calculation in my project...
Its a simple calculation But I dont know... I googled about that but can't to find the solution....
I have the following values (note:its a dynamic value)
Interest rate (per day) => 0.17%
Amount => 1500
Days => 15
How can I calculate the interest amount ????
Please helpme :(
|
If the amount is compounded every day then the total with interest after 15 days worth of interest is $$1500\times\underset{\text{15 times}}{\underbrace{1.0017\times1.0017\times\dots\times1.0017}}=1500\times1.0017^{15}
$$
(hopefully if that's not quite right you can change it so that it's the right number, i.e. compounded the right number of times)
|
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Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
|
Different approach:
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}+\frac{1}{2n+1}}{n^2}$$
$$=\sum_{n=1}^\infty\frac{H_{2n}}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}$$
where
$$\sum_{n=1}^\infty\frac{H_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}$$
$$=4\sum_{n=1}^\infty\frac{1}{2n}\left(-\int_0^1x^{2n-1}\ln(1-x)dx\right)$$
$$=-2\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{x^{2n}}{n}$$
$$=2\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}dx$$
$$=2\int_0^1\frac{\ln^2(1-x)}{x}dx+2\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$
$$=2(2\zeta(3))+2(-\frac58\zeta(3))=\frac{11}{4}\zeta(3)$$
and
$$\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{2n}dx$$
$$=\int_0^1\sum_{n=1}^\infty \frac{x^{2n}}{n^2}=\int_0^1\text{Li}_2(x^2)dx$$
$$=x\text{Li}_2(x^2)|_0^1+2\int_0^1\ln(1-x^2)dx$$
$$=\zeta(2)+2(x-1)\ln(1-x^2)|_0^1-4\int_0^1\frac{x}{1+x}dx$$
$$=\zeta(2)-4(1-\ln(2))$$
|
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Uniqueness of Unitary operator i saw the post "Polar decomposition normal operator" (Polar decomposition normal operator). There was that such a $U$ is unique iff the image of $T$ is dense. Some lines later by the comments there is that we also can say that $||T||>\delta||x||$, but this means that $T$ is invertible. Can we thus also say:
$U$ in unique iff $T$ is invertible.
I thought about that: One direction is obvious. But suppose $U$ is unitary how can we see that $T$ is not invertible? My idea was: Suppose we have that $T^*=UT$ and $T^*=U'T$, $U\neq U'$, then $(U-U')T=0$. May we say that then $T=0$ and therefore not invertible? Or is this not correct?
Thank you.
|
For a bounded operator $A$ with polar decomposition $A = UP$, $U$ is the (canonical, if you'd like) partial isometry
$$
(A^*A)^{\frac{1}{2}}h \stackrel{U}{\mapsto} Ah
$$
from $Ran(A^*A) = Ran(A^*A)^{\frac{1}{2}}$ to $Ran(A)$. If $A$ is invertible, so is $A^*A$. This make $U$ unique.
Assuming $U$ admits unitary extensions and has more than one extension, $Ran(A^*A)$ is not dense. So $A$ is not invertible.
|
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Another question about integrable functions with a transform I am an engineering student, and taking a real analysis course at demand of my advisor, my inexperience in proofs is giving me hard time. I stumbled upon this example, whose proof left as an exercise. Could you please give me some clues? Here is the example:
$f\in L^1(\mathbb R^n).$ Show that
$$
\lim_{t\rightarrow 1} \int_{\mathbb R^n}|f(x)-t^nf(tx)|\; dm=0
$$
|
Here is a lacunary solution.
*
*Approximate $f$ in $\mathbb L^1$ by a simple function (a linear combination of characteristic functions of measurable sets: this reduced the proof where $f$ is such a function.
*By linearity, do it when $f=\chi_B$, where $B$ is a Borel subset of $\mathbb R^n$, with finite Lebesgue measure.
*By outer regularity, we are reduced to threat the case $B$ open.
*Use a dominated convergence argument.
|
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Stuck on simple partial integration $$\int_0^3\frac{|x-y|}9dy=\frac19\left(x-\frac32\right)^2+\frac14$$
Could someone enlighten me regarding this partial integration?
I feel like i'm missing something but I dont know what I am doing wrong.
|
Hint
$$\int_0^3\frac{|x-y|}{9}dy=\int_0^x\frac{x-y}{9}dy+\int_x^3\frac{y-x}{9}dy$$
|
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Two different M/G/Infinity queues I need some help with this problem. We have a service system where customers arrive randomly, following a Poisson process (intensity λ). The times that the customers spend in service are independent, following a Weibull distribution with parameters α > 0 and β > 0. The times in service are also independent of the arrival process. The system has an infinite number of service stations and infinite capacity.
In the problem, the system is simulated and we get a set of data. The data is a random sample of number of customers at random times. We do two simulations with the parameters:
α1 = 0.0241
β1 = 2
α2 = 1.75
β2 = 1
Since the system has got an infinite number of service stations and an infinite capacity, we get two different M/G/∞ queues, and in this case it is known that L/W = λ, which gives me the expected number of customers in the system L = λ*W. W is the theoretic average (the expected value) of the Weibull distribution. We get
L1 = 19.980
L2 = 2.000
My question is: What's the reason to why the two simulations behave so differently?
|
The analytic result you're discussing for the mean of $L$ is
$$ \lambda \alpha \Gamma\left(1+\frac{1}{\beta}\right)$$
which I guess you've obtained, your question definitely suggests this. It's the product of $\lambda$ and the mean waiting time a single customer experiences which is the mean of the Weibull distribution where the function $\Gamma(\cdot,\cdot)$ is the incomplete Gamma function. If we set $\lambda$ to be a constant (say, 1) then we can draw a 3D plot and contour plot for different values of $\alpha$ and $\beta$ using Wolfram Alpha.
Observe how steep the function is for small values of $\alpha$ and consider where your $(\alpha,\beta)$ pairs are in this space. As the contour plot shows the function is very steep toward the left, perhaps the behaviour of the Weibull distribution mean has surprised you?
|
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How to work out this easy fraction? I need help working out this fraction, I know it seems quite easy but I'm a bit stuck.
The question is:
$$\frac{\frac {5}{2}}{\frac{5}{9}}$$
My attempt was changing the denominators by multiplying by $2$ to make $18$ and then changing the same way the nominators:
$$\frac{\frac {45}{18}}{\frac{10}{18}}$$
The thing is that I'm not sure how to simplify it as no numbers go into all of them. Thank you, I know this is quite easy.
Question: Why is it considered a division if the easiest solution is by multiplication?
|
5/2 * 9/5
Multiply by what you're dividing by's reciprocal.
|
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Solutions of $f(x)\cdot f(y)=f(x\cdot y)$ Can anyone give me a classification of the real functions of one variable such that $f(x)f(y)=f(xy)$? I have searched the web, but I haven't found any text that discusses my question. Answers and/or references to answers would be appreciated.
|
There is a classification of the functions $f:\mathbb R\to\mathbb R$ satisfying
$$
f(x+y)=f(x)+f(y), \quad\text{for all $x,y\in\mathbb R$}. \qquad (\star)
$$
These are the linear transformations of the linear space $\mathbb R$ over the field $\mathbb Q$ to itself. They are fully determined once known on a Hamel basis of this linear space (i.e., the linear space $\mathbb R$ over the field $\mathbb Q$).
This in turn provides a classification of all the functions $g:\mathbb R^+\to\mathbb R^+$ satisfying
$$
g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R^+$},
$$
as they have to be form $g(x)=\mathrm{e}^{f(\log x)}$, where $f$ satisfies $(\star)$. Note that $g(1)=1$, for all such $g$.
Next, we can achieve characterization of functions $g:\mathbb R\to\mathbb R^+$ satisfying
$$
g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R$},
$$
as $g(-x)=g(-1)g(x)$, which means that the values of $g$ at the negative numbers are determined once $g(-1)$ is known, and as $g(-1)g(-1)=g(1)=1$, it has to be $g(-1)=1$. Also, it is not hard to see that only acceptable value of $g(0)$ is $0$.
Finally, if we are looking for $g:\mathbb R\to\mathbb R$, we observe that, if $g\not\equiv 0$, and $x>0$, then $g(x)=g(\sqrt{x})g(\sqrt{x})>0$. Thus $g$ is fully determined once we specify whether $g(-1)$ is equal to $1$ or $-1$.
Note that if $g: \mathbb R\to\mathbb R$ is continuous, then either $g\equiv 0$ or $g(x)=|x|^r$ or $g(x)=|x|^r\mathrm{sgn}\, x $, for some $r>0$.
|
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|
In how many ways can we select n objects from a collection of size 2n that consists of n distinct and n identical objects? In how many ways can we select $n$ objects from a collection of size $2n$ that consists of $n$ distinct and $n$ identical objects?
The answer is $2^n$ and I really don't see how they get this. Selecting $n$ distinct from $2n$ is $\binom{2n}{n}$.
|
Let $A=\{a_1,\ldots,a_n\}$, where the $a_k$ are mutually distinguishable, and let $S$ be $A$ together with $n$ indistinguishable objects. An $n$-element subset $X$ of $S$ is completely determined when you know $X\cap A$: if $|X\cap A|=k$, the remainder of $X$ is just $n-k$ of the indistinguishable objects. $A$ has $2^n$ subsets, so there are exactly $2^n$ different possibilities for $X\cap A$ and therefore for distinguishable sets $X\subseteq S$ of cardinality $n$.
|
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|
Cups of water from a bucket We have an empty container and $n$ cups of water and $m$ empty cups. Suppose we want to find out how many ways we can add the cups of water to the bucket and remove them with the empty cups. You can use each cup once but the cups are unique.
The question: In how many ways can you perform this operation.
Example: Let's take $n = 3$ and $m = 2$.
For the first step we can only add water to the bucket so we have 3 choices.
For the second step we can both add another cup or remove a cup of water.
So for the first 2 steps we have $3\times(5-1) = 12$ possibilities.
For the third step it gets more difficult because this step depends on the previous step.
There are two scenarios after the second step. 1: The bucket either contains 2 cups of water or 2: The bucket contains no water at all.
1) We can both add or subtract a cup of water
2) We have to add a cup of water
So after step 3 we have $3\times(2\cdot3 + 2\cdot2) = 30$ combinations.
etc.
I hope I stated this question clearly enough since this is my first post. This is not a homework assignment just personal curiosity.
|
Let $L=n+m$ and $D=n-m \ge 0$
Hint 1: consider a sequence $X=(x_1,x_2, ... x_L)$, associate a filled cup with $x_i=1$ and an empty with $x_i=-1$. Let $C(n,m)$ count all such binary sequences of length $L$ with the two restrictions: $\sum_{k=1}^j x_k\ge 0$, $\forall j$ and $\sum_{k=1}^L x_k =D$
Then, the total number of ways is $C(n,m) n! m!$
Hint 2: To obtain $C(n,m)$ is not trivial, but random walks counting procedures (mirror principle) might help (eg).
Edited: using the reflection principle:
Consider first counting all the paths going from $y(0)=0$ to $y(L)=D$ such that $y(t+1)=y(t) \pm 1$, and with $D\ge0$. (Recall that $L=n+m$, $D=n-m$). This unrestricted count is $C^{[u]}={n+m \choose n}={L \choose (L+D)/2}$
Now, consider the "prohibited" paths: these correspond to those that touch the $y(t)=-1$ line. By the mirror principle, these correspond one-to-one to the unrestricted paths that start from $y(0)=-2$, and this count is given by $C^{[p]}={L \choose (L+D+2)/2}={n+m \choose n+1}$
Hence, the number of allowed paths is
$$C(n,m) = C^{[u]}-C^{[p]}={n+m \choose n} - {n+m \choose n+1}=\frac{1+n-m}{n+1}{n+m \choose n} = \frac{1+D}{n+1}{L \choose n} $$
And, finally, the number of ways is
$ \frac{1+D}{n+1} \,L!$
See also the Ballot problem.
|
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|
How to find $f(x)$ and $g(x)$ when only given $f(g(x))$ I've learned how to find $f(g(x))$ when given the two $f(x)$ and $g(x)$ functions fairly easily, but I haven't found anywhere online showing how to do the opposite. For this question I'm working on I'm asked to find $f(x)$ and $g(x)$ if $\cos^2(x) = f(g(x))$.
Can anyone please help me figure out how to solve this? Also if someone could provide a website that helps explain this that would be greatly appreciated.
|
This isn't possible to do uniquely, since for example
$$
f\left(x\right)=x
$$
and
$$
g\left(x\right)=\cos^{2}x
$$
gives you the desired result. However, I think the answer they are looking for is
$$
f\left(x\right)=x^{2}
$$
and
$$
g\left(x\right)=\cos x
$$
so that
$$
f\left(g\left(x\right)\right)=\left(\cos x\right)^{2}\equiv\cos^{2}x.
$$
Sounds like you have yourself a bad teacher/textbook.
|
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|
If D is an Integral Domain and has finite characteristic p, prove p is prime. So the question is simply.
If $D$ is an integral domain and has finite characteristic prove that the characteristic of $D$ is a prime number.
This is my proof.
Assume $p$ is the characteristic of $D$. Let $a$ be a non zero element of $D$. Seeking a contradiction assume $p$ is not prime.
Then $p$ can be written as a factor: $rs=p$ for some $r$ and some $s$.
By definition $pa=0$, so $(rs)a=0$. We know that $r,s$ are non-zero, so by definition of integral domain the only way this equation can equal zero is if $a=0$ however this is a contradiction as we chose a to be a non-zero element of $D$. Therefore $p$ is a prime.
Is this proof correct? The answer I have for this problem is slightly longer and I thought I might have missed something in my proof.
|
Hint $\ $ The finite characteristic $\,n\,$ is just the size of the natural image of $\Bbb Z$ in $D\,$, via $\,1_\Bbb Z \mapsto1_D.$ This image is a subring of $D$ isomorphic to $\,\Bbb Z/n,\,$ which is a domain $\iff n\,$ is prime.
|
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|
Linear combination and Basis
*
*Consider a set of five arbitrary 2x2 matrices. Can you always write one as a linear combination of the others? Explain. Repeat for five arbitrary 3x3 matrices.
*For each of the following sets explain whether or not the set is/could be a basis for the space mentioned.
a) Four vectors in that form a loop when end-to-end connected.
...............
Problem is i didnt understand the questions.
For the first question: i know what linear combination is, but Do i have to find that five matrices? which can be linear combination of one another ? or is asking that whether any 5 matrices can be linear combination of one another or cant?
For the second question:
i have learned Standard basis of matrix, (actually learning) but there is no standard. Im just started learning and bit confused
if any1 explain it little bit i will figure it out.
Thanks
|
The first part of the first question is asking whether every set of five $2\times 2$ contains one that is a linear combination of the other four; the second part is asking the same question about sets of five $3\times 3$ matrices. The first part can be paraphrased as follows: is every set of five $2\times 2$ matrices linearly dependent? Similarly for the second part of the first question.
The second question is asking whether four geometric vectors that form a loop can be a basis. An example of such a loop in $\Bbb R^2$ is shown here:
HINT: What’s the sum of four vectors that form a loop?
|
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Find The Eigenvalues and Eigenvectors of the Hermitian Matrix Find the eigenvalues and eigenvectors of the $2\times2$ hermitian matrix.
$$\pmatrix{\epsilon_1&|V|e^{i\alpha}\\
|V|e^{-i\alpha}&\epsilon_2}$$
I know to find eigenvalues, you use $|A-\lambda I|$, but this is giving me difficult results to find an exact value for $\lambda$.
$V$, $\epsilon_1$, $\epsilon_2$, $\alpha$ are all constants.
|
We can start off by solving the more general case system in order to simplify matters:
$$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$
This produces the eigenvalue / eigenvector pairs:
*
*$\lambda_1 = \dfrac{1}{2} \left(-\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$
*$v_1 = \left(\dfrac{-(-a+d+\sqrt{a^2+4 b c-2 a d+d^2})}{2 c}, 1\right)$
*$\lambda_2 = \dfrac{1}{2} \left(\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$
*$v_2 = \left(\dfrac{-(-a+d-\sqrt{a^2+4 b c-2 a d+d^2})}{2 c}, 1\right)$
We can now use this result to write the eigenvalues and eigenvectors of the original system:
$$\begin{bmatrix}\epsilon_1&|V|e^{i\alpha} \\ |V|e^{-i\alpha}&\epsilon_2 \end{bmatrix}$$
$~~$
*
*$\lambda_1 = \dfrac{1}{2} \left(-\sqrt{\epsilon_1^2-2 \epsilon_1 \epsilon_2+4 |V|^2 +\epsilon_2^2}+\epsilon_1+\epsilon_2\right)$
*$v_1 = \left(\dfrac{-e^{i~\alpha}(-\epsilon_1+\epsilon_2+\sqrt{\epsilon_1^2+4 |V|^2 -2 \epsilon_1 \epsilon_2+\epsilon_2^2})}{2 |V|}, 1\right)$
*$\lambda_2 = \dfrac{1}{2} \left(\sqrt{\epsilon_1^2-2 \epsilon_1 \epsilon_2+4 |V|^2+\epsilon_2^2}+\epsilon_1+\epsilon_2\right)$
*$v_2 = \left(\dfrac{-e^{i~\alpha}(-\epsilon_1+\epsilon_2-\sqrt{\epsilon_1^2+4 |V|^2 -2 \epsilon_1 \epsilon_2+\epsilon_2^2})}{2 |V|}, 1\right)$
|
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|
Method of characteristics: $xu_t -tu_x = u$ I am just learning the method of characteristics. Suppose I want to solve
$$xu_t - tu_x = u \quad u(x,0) = h(x)$$
I write
$$\dot{t}(s) = x\\
\dot{x}(s) = -t \\
\dot{z}(s) = z$$
If $t(0) = 0$ and $x(0) = x_0$, we have (I think)
$$t(s) = x_0\sin(s)\\
x(s) = x_0\cos(s)\\
z(s) = h(x_0)e^s$$
Where do I go from here? First I'd like to get rid of the $s$, then I'd like to get rid of the $x_0$, but I'm not seeing how to do that.
|
Clearly, the domain of a solution of $xu_t-tu_x=u$, can not contain any circle centered at the origin. But this is the only restriction!
Take $\Omega=\mathbb R^2\smallsetminus C$, where $C$ is a curve starting from the origin and going to infinity, assume also that $C$ does not intersect to positive $x$-axis. Let $h=h(x), x\in(0,\infty)$ be a $C^1$ function. Then using the ideas of the previously exposed solution, you can obtain a unique solution $u=u(x,t)$ of the PDE defined in $\Omega$, such that $u(x,0)=h(x)$.
|
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|
Find $c_1,c_2,c_3\in\mathbb{Q}$ such that $(1+\alpha^4)^{-1}=c_1+c_2\alpha+c_3\alpha^2$ in $\Bbb Q(\alpha)$. Let $\alpha\in \overline{\mathbb{Q}}$ be a root of $X^3+X+1\in\mathbb{Q}[X]$. So this is the minimal polynomial of $\alpha$ because it's irreducible in $\mathbb{Q}[X]$. I had to find the minimal polynomials of $\alpha^{-1}$ and $\alpha-1$. After that I have to do what the title says. I found the minimal polynomials $f_1=X^3+3X^2+4X+4$ and $f_2=X^3+X^2+1$ of $\alpha^{-1}$ and $\alpha-1$, respectively. Now I suppose I have to use this result to solve the question, but I don't have any idea how to this. I need help. Thanks.
|
Hint
Apply the extended Euclidean algorithm to find polynomals $g,h$ with
$$
g(X^4 + 1) + h(X^3 + X + 1) = 1.
$$
Then consider this equation mod $X^3 + X + 1$ and plug in $\alpha$.
Note: It's not strictly neccesary, but you may want to reduce $X^4 + 1$ mod $X^3 + X + 1$ first.
|
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|
solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you please explain me the solution?
|
We have $$(x-1)^2=x^2+1-2x\ge0\iff x^2+1\ge 2x$$
and since $\sin^2\theta\ge0$ hence we have $x>0$ and then
$$1\ge\sin^2\theta =\frac{x^2+y^2+1}{2x}\ge1\quad (>1\;\text{if}\; y\ne 0\;\text{which gives a contradiction})$$
hence $y=0$ and
$$\sin^2\theta=\frac{x^2+1}{2x}=1\iff x=1$$
|
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|
LCR Odds - 6 'C's in a row. Played the game LCR. One unlucky guy rolled 3 - 'C's and then to make matters worse rolled 3 - 'C's again on his next turn. So what are the odds of that happening?
|
While @Ross's answer is entirely correct, there is a broader question here. The question asked "what are the odds of that happening". What is "that"? The specific event of rolling six C's in a row does indeed have probability $\frac{1}{46656}$.
However, if instead your unlucky friend had rolled six R's in a row, you would think that was equally unlikely, and you would still be here asking that same question. The probability of rolling either six C's or six R's in a row is twice Ross's answer, or $\frac{2}{46656}=\frac{1}{23328}$.
Another unlucky situation is losing three counters (the maximum), for two turns in a row. The probability of that happening is $(\frac{1}{2})^6=\frac{1}{64}$, which is actually not that unlikely at all.
|
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|
Converting Context Free Grammar to Chomsky Normal Form This is an exercise that I had to complete in my class and I struggled a lot with it
$$\begin{align*}
&S\to 0A0\mid 1B1\mid BB\\
&A\to C\\
&B\to S\mid A\\
&C\to S\mid\epsilon
\end{align*}$$
Added from comments: This is what I have but I am sure it is wrong: replace epsilon productions:
$$\begin{align*}
&S\to 0A0\mid 1B1\mid BB\\
&A\to C\\
&B\to S\mid A\\
&C\to S
\end{align*}$$
replace unit productions:
$$\begin{align*}
&S\to 0A0\mid 1B1\mid BB\\
&A\to C\mid 0A0\mid 1B1\mid BB\\
&B\to S\mid A\\
&C\to S\mid 0A0\mid 1B1\mid BB
\end{align*}$$
replace mixed productions:
$$\begin{align*}
&S\to T0AT0\mid T1BT1\mid BB\\
&A\to C\mid T0AT0\mid T1BT1\mid BB\\
&B\to S\mid A\\
&C\to S\mid T0AT0\mid T1BT1\mid BB
\end{align*}$$
|
(a) Eliminate -productions.
Since C can produce , A can also produce , so B can produce , so S can produce .
therefore, we need to change the productions for every symbol.
S → 00 | 0A0 | 11 | 1B1 | B | BB
A → C
B → S | A
C → S
(b) Eliminate any unit productions in the resulting grammar.
We note that A, B, and C will all just produce S again. So we remove them.
S → 00 | 0A0 | 1B1 | BB
A → 00 | 0A0 | 1B1 | BB
B → 00 | 0A0 | 1B1 | BB
C → 00 | 0A0 | 1B1 | BB
(c) Eliminate any useless symbols in the resulting grammar.
The variable C has now become unreachable. We also remove A and B, because they
are exactly equal to S. Our grammar becomes:
S → 00 | 0S0 | 1S1 | SS
(d) Put the resulting grammar into Chomsky Normal Form.
To make this a CNF grammar, we first create variables A → 0 and B → 1. We then
divide the two productions of length 3 using variables C and D.
Our final CNF grammar is:
S → AA | AC | BD | SS
A → 0
B → 1
C → SA
D → SB
|
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|
Show that $\sum_{k=2012}^{n} 2^k\binom{n}{k} = \Theta(3^n)$ In this question we are asked to show that
$\sum_{k=2012}^{n} 2^k\binom{n}{k} = \Theta(3^n)$
What I did:
$\sum_{k=2012}^{n} 2^k\binom{n}{k} = \sum_{k=2012}^{n} 2^k*1^{n-k}\binom{n}{k} \leq \sum_{k=0}^{n} 2^k*1^{n-k}\binom{n}{k} = (2+1)^n = 3^n$, using newton's binomial.
So obviously, $\sum_{k=2012}^{n} 2^k\binom{n}{k} = O(3^n)$
How do I show that $\sum_{k=2012}^{n} 2^k\binom{n}{k} = \Omega(3^n)$? Could anyone please point me in the right direction?
|
You note that the term you are omitting from the Newton binomial are all smaller than $\binom{n}{k}$ in growth (the $2^k$ are smaller than $2^{2012},$ so constant. And the binomial is at most a polynomial of degree $2012,$ so subexponential.
|
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|
A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is:
$A. 1/2$
$B. 3/5$
$C. 1/5$
$D. 2/3$
What I did: The probability should be $(2/3)*(1/3)+(2/3)*(2/3)*(2/3)*(1/3)+(2/3)*(2/3)(2/3)*(2/3)*(2/3)*(1/3)+...$, which should be $2/5$.
But this answer doesn't match with the options. Where did I go wrong?
|
You didn't go wrong. The probability of "success" is 1/3, so the probability of succeding in $2k$ tosses is
$$\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}$$
And
$$\sum_{k=1}^\infty\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}=\frac{1}{3} \frac{3}{2}\sum_{k=1}^\infty\left(\frac{4}{9}\right)^k =\frac{1}{2} \frac{4/9}{1-4/9}=\frac{2}{5}$$
|
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|
find the limit of $\mathop {\lim }\limits_{x \to 0 } \frac{a}{x}\left[ {\frac{x}{b}} \right]$ $$a,b \gt 0$$
$$\mathop {\lim }\limits_{x \to 0 } \frac{a}{x}\left[ {\frac{x}{b}} \right]$$
So, I know that if x is $x \in \mathbb{Z}$ then the limit is $a\over [b]$
I couldn't figure out the solution for $x \notin \mathbb{Z}$
By the way, $[x]$ notation is equal to floor(x).
I'll be glad for a direction here
|
Hint
Assume that $b>0$, prove that
$$\mathop {\lim }\limits_{x \to 0^+ } \frac{a}{x}\left[ {\frac{x}{b}} \right]=0$$
and
$$\mathop {\lim }\limits_{x \to 0^{-} } \frac{a}{x}\left[ {\frac{x}{b}} \right]=\infty$$
|
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Why is true? $\mathop {\lim }\limits_{x \to 0} \frac{x}{a}\left[ {\frac{b}{x}} \right] = \frac{b}{a}$ $$\begin{array}{l}a,b > 0\\\mathop {\lim }\limits_{x \to 0} \frac{x}{a}\left[ {\frac{b}{x}} \right] = \frac{b}{a}\\\end{array}
$$
I asked already a similar question, but I'm still not sure what makes it true.
As $x$ is decreasing to $0$, $x \over a$ is converging to $0$ but $\left[ {\frac{b}{x}} \right]$ is converging to $\infty$, So we left with $0*\infty$ which isn't helpful.
How can you solve it?
|
We have,
$\displaystyle\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\left[\dfrac{b}{x}\right]}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\left(\dfrac{b}{x}-\left\{\dfrac{b}{x}\right\}\right)}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\cdot\dfrac{b}{x}-\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{b}{a}-\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
$\displaystyle=\mathtt{\dfrac{b}{a}-\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
The limit $\,\,\displaystyle\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}=0$
because the term $\dfrac{x}{a}$ is $0$ , if we put $x=0$ and the term $\left\{\dfrac{b}{x}\right\}\in[0,1)\,\,\,\,\,\because\{y\}\in[0,1)$
So, $0\times[0,1)=0$
Hence, the required limit is $\dfrac{b}{a}$
|
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Derive the Frenet equations I was looking for a derivation of the Frenet equations. I've been following this reference but I've been having problems in understanding this statement (found before Eq.$(2.19)$):
When r'($s+\Delta s$) is moved from $Q$ to $P$, then r'($s$), r'($s+\Delta s$) and r'($s+\Delta s$)-r'($s$) form an isosceles triangle, since r'($s+\Delta s$) and r'($s$) are unit tangent vectors. Thus we have $|\textbf{r}'(s+\Delta s)-\textbf{r}'(s)|=\Delta \theta \cdot 1=\Delta \theta=|\textbf{r}''(s)\Delta s| $ as $\Delta s \to 0$ [...]
My problem is in understanding the last string: I mean how can I prove mathematically that $|\textbf{r}'(s+\Delta s)-\textbf{r}'(s)|=\Delta \theta \cdot 1=\Delta \theta$. Is this taken from the dot product propriety:
$$|\textbf{a}-\textbf{b}|^2=a^2+b^2 + 2 ab \cos(\theta),$$
assuming $\textbf{a}$ and $\textbf{b}$ are unitary vectors and taking the angle between the vectors approximately zero, thus Taylor expanding the cosine to second order?
|
You've mis-typed things a little -- in each place where you have $|r'(s + \Delta s) - r'(s + \Delta s)|$, you should have $|r'(s + \Delta s) - r'(s)|$; I suspect this is a cut-and-paste error.
The claim the authors make -- that the difference vector EQUALS $\Delta \theta \cdot 1$ -- is false. But what's true is that if you have an arc of a unit circle subtending angle $\Delta \theta$, then the length of the arc is also $\Delta \theta$, so that the rightmost arrow in the Figure 2.4 that you refer to (the one pointing SE) has a length that's APPROXIMATELY the same as the length of a circle arc between its two endpoints. As $\Delta \theta$ gets small, this approximation gets better and better (basically, it's the approximation $\sin x \approx x$ for small $x$), so the rest of the argument is valid.
I have to say, a quick look at that page doesn't give me a lot of confidence. You might want to look at Barrett O'Neill's book, "Elementary Differential Geometry", or Millman and Parker's book, which is a bit more sophisticated, or doCarmo, which sits somewhere in the middle.
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|
probability of drawing balls from an urn An urn contains four balls: one red, one green, one yellow, and one white.
Two balls are drawn without replacement from the urn. What is the probability of getting a red ball and a white ball? Assume that the balls are
equally likely to be drawn
Here's what I've tried:
Probability of a red or a white ball on first draw: $\frac{1}{4}$
Probability of a red or a white ball on second draw: $\frac{1}{3}$
Total Probability : $\frac{1}{4}\cdot\frac{1}{3}=.0833$, but the correct answer is $0.167$
|
Probability of getting a red or white ball on first draw:$\frac{1}{2}$
probability of getting the remaining ball that is either red or white $\frac{1}{3}$
probability both of these happen: $\frac{1}{2}\cdot \frac{1}{3}\approx0.167$
|
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In differential equations, do transient state and steady state always go hand in hand? Do transient state and steady state go hand in hand in differential equations? Meaning, if there is a transient state, is there always going to be a steady state?
Also, if there are neither of them, then is it always pure resonance?
Could a DE not have any of those states?
Please explain and answer in the simplest terms. Thank you
|
Lets say we have a spring with an external force.
We solve the second order ODE and arrive at:
$$x(t) = \dfrac{e^{-4 t}}{9} [3 \cos(4 \sqrt{3} t) - 11 \sqrt{3}) \sin(4 \sqrt{3} t)] + 2 \sin(8 t) \\ y(t) = 2 \sin(8t)$$
A plot of these functions is:
You can see that the $e^{-4t}$ becomes negligible when $t$ is large. The steady state solution is shown in blue and the steady-state plus transient is shown in red.
These transient terms in the solution, when they are significant, are sometimes called the transient solutions. In many physical problems, the transient solution is the least important part. However, there are cases where it is of major importance.
When the transient terms are negligible, only the $2\sin(8t)$ remains. This is called the steady-state term or steady-state solution, since it indicates the behavior of the system when conditions have become steady. You can see in the graph, that the steady-state solution is periodic and has the same period as that of the applied external force (in this case, that was, in this example, $F(t) = 24 \cos(8t)$).
Electrical circuits with resistors, capacitors and inductors (like an RLC circuit) look at transient and steady state quite a bit, so you might want to explore those.
Look at the book by Tenenbaum and Pollard for more examples regarding the damped and undamped frequency and why these things can become critically important.
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Proof about GCD's Prove that if $a, b$ and $c$ are integers with $b \neq 0$ and $a=bx+cy$ for some integers $x$ and $y$, then $\text{gcd}(b,c) \le \text{gcd}(a,b).$
I don't understand how to show (b,c) is less than (a,b)?
|
$\ \begin{eqnarray}\color{}{(b,c)}&&\mid\, \color{sienna}b,\ \ \ \color{sienna}c \\ \Rightarrow\ (b,c)&&\mid \,\color{sienna}b\,x\!\!+\!\!\color{sienna}c\,y=\color{#c00}a\!\end{eqnarray}\bigg\rbrace\ $ so $\ \bigg\lbrace\begin{eqnarray}(b,c)&\mid&\ \color{#c00}a,\,b\\ \,\Rightarrow (b,c) &\color{#0a0}\le& (a,b) \end{eqnarray}\bigg\rbrace\ $ by $\ \bigg\lbrace\, \begin{eqnarray} &&c\rm ommon\ divisor \\ \color{#0a0}\le &&{\it greatest}\rm\,\ common\ divisor\end{eqnarray}$
|
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For every $n \in \mathbb{N}$, find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times. For every $n \in \mathbb{N}$, (in this problem $\mathbb{N}$ starts at $n = 1$) find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times.
The function I chose is the following:
$$
f_n(x) = \begin{cases} x^{n+1} &\mbox{if } x \le 0\\
x^{n} &\mbox{if } x > 0 \end{cases}
$$
Now I can't think of the best way to prove that $f$ is differentiable $n$ times at $0$ but not $n+1$ times.
|
Consider $x^\alpha\sin(x^{-1})$ for a suitable $\alpha$.
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find $a_n$, $a_n=\sqrt{\frac{a_{n-1}^2+a_{n+1}^2}{2}}, a_1=10(a_n\in \mathbb{N},n=2,3,4,\cdots)$ I would appreciate if somebody could help me with the following problem
Q: find $a_n$
$$a_n=\sqrt{\frac{a_{n-1}^2+a_{n+1}^2}{2}}, a_1=10(a_n\in \mathbb{N},n=2,3,4,\cdots)$$
|
Follow Ewan's answer.
Obviously, $c_2\ge0$, otherwise $b_n=a_n^2=c_1+c_2n<0$ for large $n$.
If $c_2>0$, then $a_{n+1}-a_{n}>0$.
However, $a_{n+1}-a_{n}=\sqrt{c_1 + c_2 n+c_2}-\sqrt{c_1 + c_2 n}=\frac{c_2}{\sqrt{c_1 + c_2 n}+\sqrt{c_1 + c_2 n+c_2}}<\frac{c_2}{2\sqrt{c_1+c_2n}}$
So for $n>c_2-c_1/c_2$, $0<a_{n+1}-a_n<\frac{c_2}{2\sqrt{c_2^2}}=1/2$, which contradicts the condition that $a_n$ and $a_{n+1}$ are both natural number.
So $c_2=0$, and $c_1=a_1-0*1=10$,
In conclusion $a_n=c_1+0*n=a_1=10$.
|
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how to show $E[|X|]= \sigma$ where $X \sim N(0, \sigma^2)$ let $X \sim N(0, \sigma^2)$ I want to show $$E[|X|]= \sigma$$
thanks for help
|
Let $Z\sim N(0,1)$ so that $X=\mu+\sigma Z\sim N(\mu,\sigma^2)$.
Then $Y=|X|$ has the Folded Normal Distribution if $\mu\ne 0$ and the Half-Normal Distribution when $\mu=0$.
In your case $Y$ has the Half-Normal Distribution with cdf $F$ and pdf $f$
$$
\begin{align}
F(y) & = 2 \Phi\left(\frac{y}{\sigma}\right) - 1 = \int_0^y \frac{1}{\sigma} \sqrt{\frac{2}{\pi}} \exp\left(-\frac{x^2}{2 \sigma^2}\right) \, dx, \quad y \in [0, \infty) \\
f(y) &= \frac{2}{\sigma} \phi\left(\frac{y}{\sigma}\right) = \frac{1}{\sigma} \sqrt{\frac{2}{\pi}} \exp\left(-\frac{y^2}{2 \sigma^2}\right), \quad y \in [0, \infty)
\end{align}
$$
The even order moments of $Y$ are the same as the even order moments of $\sigma Z$:
$$
\Bbb E(Y^{2n}) = \sigma^{2n}\Bbb E(Z^{2n}) = \sigma^{2n} \frac{(2n)!}{n! 2^n}
$$
using the $n$-th derivative of the mgf $M_Z(t)=e^{t^2/2}$ in $t=0$.
For the odd order moments we have by substituting $y^2/2\sigma^2=z$:
$$
\Bbb E(Y^{2n+1}) = \int_0^\infty y^{2n+1} \sqrt{\tfrac{2}{\pi \sigma^2}} e^{-y^2/2} \, dy = \sigma^{2n+1} 2^n \sqrt{\tfrac{2}{\pi}}\underbrace{\int_0^\infty z^n e^{-z} \, dz}_{\Gamma(n+1)=n!} = \sigma^{2n+1} 2^n \sqrt{\tfrac{2}{\pi}} n!
$$
In particular, we have $\Bbb E(Y) = \sigma \sqrt{2/\pi}$ and $\text{Var}(Y) = \sigma^2(1 - 2 / \pi)$.
|
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Relation between Supp and sheaves Let $\mathcal{F}$ be a cohorent sheaf on projective scheme $X$.
My question is simple... If $\operatorname{dim}\operatorname{Supp}\mathcal{F}$ is zero, then $\mathcal{F}(n) =\mathcal{F}$ for any integer $n$??
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One should read the comments by Matt E before reading the following. I am keeping this just as a record.
Let $X = \hbox{Proj} \, (K [x,y])$ and $Y = \hbox{Proj} \, (K[x,y]/(x^2))$. Write $i: Y \hookrightarrow X$ and $F = i_* \mathcal{O}_Y$. Then $\dim \hbox{Supp} \, F = 0$. One has $\Gamma(X, F) = 1$, but $\Gamma(X, F(i)) = 2$ for $i \ge 1$.
I belive that this has to do with Hilbert-polynomial. Since $\dim F = 0$, the degree of the Hilbert polynomial is $0$, i.e., it is a constant. Since degree of $F$ is $2$, one has $\Gamma(X, F(i)) = 2$ for $i \gg 0$. However, this does not tell us the behavior of the Hilbert-funtion when $i$ is "small".
In the above $\Gamma(X, F) = 1$ is incorrect. The exact sequence of graded $S= K[x,y]$-modules
$$
0 \to (x^2) \to S \to S/(x^2) \to 0
$$ induces an exact sequence of $\mathcal{O}_X$-modules
$$
0 \to \mathcal{O}_X (-2) \to \mathcal{O}_X \to F \to 0.
$$
Taking $\Gamma(X,-)$, we obtain an exact sequence
$$
H^0(X, \mathcal{O}_X (-2)) \to H^0(X, \mathcal{O}_X) \to H^0(X, F) \to H^1(X, \mathcal{O}_X (-2)) \to H^1(X, \mathcal{O}_X ).
$$
Since $H^0(X, \mathcal{O}_X (-2)) = H^1(X, \mathcal{O}_X ) = 0$ and $H^0(X, \mathcal{O}_X) = H^1(X, \mathcal{O}_X (-2)) = 1$, we see that $H^0(X, F) = \Gamma(X,F) = 2$.
|
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$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^TA$ is diagonalisable? I have been set some work to do over the holidays, and one of the questions gives a hint that is as follows:
$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^TA\text{ is diagonalisable}$.
I know that
*
*$A\in\mathrm{M}_{n\times n}(\mathbb{R})\implies A^TA\text{ is diagonalisable}$
*$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^*A\text{ is diagonalisable}$
where the former can be thought of as a particular case of the latter. Both those statements are true because $A^*A$ is self adjoint, and we can then apply the Spectral Theorem for normal operators.
But is the statement at the top of my question true, or has the lecturer simply mistyped one of the two facts that I've written? If it is true, I can't see how to prove it, so any hints would be appreciated.
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This is false.
$$
A=\pmatrix{\frac{1}{2}+i&1\\-1-\frac{i}{2}&i}\qquad A^TA=\pmatrix{2i&1\\1&0}\qquad \mathrm{Spectrum}(A^TA)=\{i\}
$$
|
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|
Examples where derivatives are used (outside of math classes) I want to know what is the use of derivatives in our daily life.
I have searched it on google but i haven't find any accurate answer. I think it is mostly used in Maths but I want to know its use in other departments i.e physics, chemistry, biology and economics.
|
All of the above.
It is actually easier to explain physics, chemistry, economonics, etc with calculus than without it.
For example:
Velocity is derivative of position with time.
Derivative of momentum (by time) is force.
Derivative of Gibbs free energy with number of atoms is chemical potential.
Etc.
I do honestly use calculus daily and I am not a mathematician. Please feel free to add to this list.
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Let T = {1000,1001...,9999}. How many numbers have at least one digit that is 0... Let T = {1000,1001...,9999}. How many numbers have at least one digit that is 0, at least one that is 1 and at least one that is 2. For example, 1072 and 2101 are two such numbers.
I have no idea to solve such probability question, I should use Permutation or Combination? why the answer is 150.
Thanks you guys.
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Let's break up the different cases, shall we?
*
*If the leading digit $d$ is not one of the special digits, that leaves us with $d \in \left\{3,4,5,6,7,8,9\right\}$. So we have $7$ different options for the leading digit. After that, there are $3! = 6$ different options for our special digits — $\left\{(d,0,1,2),(d,0,2,1), (d,1,0,2),(d,1,2,0),(d,2,0,1),(d,2,1,0)\right\}$.
So the total number of combinations is $7 \cdot 3! = 42$.
*If the leading digit is a special digit $s$, it has to be either $1$ or $2$. It can't be $0$, as that would bring the number below $1000$, which is not allowed.
*
*If the free digit $f$ is not equal to one of the remaining special digits, there are $3!$ ways of placing those digits, with $8$ options for the free digit — $f \in \left\{s,3,4,5,6,7,8,9\right\}$. This gives us $3! \cdot 8 = 48$ different options.
*If $f$ is equal to one of the remaining special digits, there are $3$ ways of placing the other remaining digit and there are of course $2$ special digits that $f$ can be equal to, giving us $2\cdot3=6$ options.
As noted earlier, there are $2$ possible leading special digits, giving us $2\cdot(48+6) = 108$ options for a leading special digit.
Adding these two cases gives us $42 + 108 = 150$, which is the answer you had.
|
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Lemma 1.3.4(b) in Bruns and Herzog My question refers to the proof of the second of the following lemma given in Cohen-Macaulay rings by Bruns and Herzog.
Lemma 1.3.4 (Bruns and Herzog): Let $(R,\mathfrak m,k)$ be a local ring, and $\phi:F \rightarrow G$ a homomorphism of finite $R$-modules. Suppose that $F$ is free, and let $M$ be an $R$-module with $m \in \operatorname{Ass}(M)$. Suppose that $\phi \otimes M$ is injective. Then
(a) $\phi \otimes k$ is injective
(b) if $G$ is a free $R$-module, then $\phi$ is injective, and $\phi(F)$ is a free direct summand of $G$.
I can see why (a) is true. But for (b), Bruns and Herzog write: "one notes that its conclusion is equivalent to the injectivity of $\phi \otimes k$. This is an easy consequence of Nakayama's lemma."
Question 1: I find this statement confusing, since if the conclusion of (b) is equivalent to the injectivity of $\phi \otimes k$, which we already proved in (a), then why do we need the extra assumption that $G$ is free?
Question 2: What would be a proof of (b)?
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Let $(R,\mathfrak m,k)$ be a local ring, and $\phi:F \rightarrow G$ a homomorphism of finite free $R$-modules. Then $\phi \otimes k$ is injective iff $\phi$ is injective, and $\phi(F)$ is a free direct summand of $G$.
"$\Leftarrow$" This is not difficult and I leave it to you.
"$\Rightarrow$" Let $\{e_1,\dots,e_m\}\subset F$ be an $R$-basis. (In the following we denote $\phi \otimes k$ by $\overline{\phi}$, and $G\otimes k$ by $\overline G$.) Then $\overline{\phi}(\overline e_1),\dots,\overline{\phi}(\overline e_m)$ are linearly independent over $k$ and we can find $\{\overline g_{m+1},\dots,\overline g_n\}\subset\overline G$ such that $\{\overline{\phi}(\overline e_1),\dots,\overline{\phi}(\overline e_m),\overline g_{m+1},\dots,\overline g_n\}$ is a $k$-basis of $\overline G$. Nakayama's Lemma shows that $\{\phi(e_1),\dots,\phi(e_m),g_{m+1},\dots,g_n\}$ is a minimal system of generators for $G$. Since $\hbox{rank}\ G=\dim_k\overline G=n$, it follows that $\{\phi(e_1),\dots,\phi(e_m),g_{m+1},\dots,g_n\}$ is an $R$-basis of $G$. This shows that $G=\phi(F)\oplus\langle g_{m+1},\dots,g_n\rangle$ and $\phi $ injective.
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Why does $\left(1+ 1/k\right)^k$ converge to $e$ as $k \to\infty$? I came across this when learning about sequences and series. No one proved it to me tho! Is there a link anywhere? Or would it be beyond what I know? (First term of analysis)
|
Of course, this depends on how you define $e$. Here is something that would convince a calculus student.
Taking logs we have:
$\log((1 + \frac{1}{k})^k) = k \log(1 + \frac{1}{k})$
So we can take this limit as $k\rightarrow\infty$ using l'hospitals rule.
$$ \lim_{k\rightarrow\infty} k \log(1 + \frac{1}{k}) = \lim_{k\rightarrow\infty} \frac{\log(1 + \frac{1}{k})}{\frac{1}{k}} = \lim_{k\rightarrow\infty} \frac{-\frac{1}{k^2 + k}}{-\frac{1}{k^2}} = \lim_{k\rightarrow\infty} \frac{k^2}{k^2 + k} = 1$$
Hence, using continuity of the function $e^x$, we conclude that
$$\lim_{k\rightarrow\infty} (1 + \frac{1}{k})^k = e^1 = e$$
|
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Initial Value Problem
We have initial value problem $y''=x^{1/3}y$ with $y(0)=y'(0)=0$.
Does it have a unique solution?
I have tried using the Picard-Lindelöf theorem, but I cannot reduce it to a order 1 ODE. Thank you.
|
Check this
Theorem: Let $p(t), q(t)$, and $g(t)$ be continuous on $[a,b]$, then the differential equation
$$y'' + p(t)y' + q(t)y = g(t),\quad y(t_0) = y_0,\, y'(t_0) = y'_0 $$
has a unique solution defined for all $t$ in $[a,b]$
|
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Solve $x^2+x+3=0$ mod $27$ I was preparing for my Number Theory class for next semester and one of the questions that I came upon is to solve $x^2+x+3=0$ mod $27$. I have seen modular arithmetic before but never one that involved using mod with an equation. I plugged it into WolframAlpha and it gave me $x=11,15$. My question is, how did wolfram alpha solve this? Is there some kind of procedure? My initial try was to use Brute Force and plug in every value possible until it is divisible by 27 but I would like to see how it is solved.
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There is no harm in completing the square. We have that $2^{-1}=-13$, so $x^2+x+3=x^2-26x+3=0$ gives $(x-13)^2+3-13^2=0$, or $(x-13)^2=4=2^2$. This means that $(x-13)^2-2^2=0$ so that $(x-15)(x-11)=0$. Note you cannot really conclude $x-15=0$ or $x-11=0$ right away, since we're not working over an integral domain.
ADD Let $p$ be an odd prime, $p\not\mid a$. Then $x^2=a\mod p^k$ has at most two (incongruent) solutions.
P Suppose $x^2\equiv x_0^2\mod p^k$. Then $p^k\mid (x-x_0)(x+x_0)$. We cannot have $p$ dividing both $x-x_0,x+x_0$ for this means say $p\mid 2x_0$, which is impossible. Thus $x= \pm x_0\mod p^k$.
|
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Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after long evaluations and considering many variations of product of polylogarithm integrals.
I think there is an easier approach to get the solution, any ideas ?
|
I think it is reasonable to start with:
$$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=\sum_{k=1}^{+\infty}\frac{H_k}{k^4}+\sum_{k=1}^{+\infty}\frac{H_k}{k^2}\sum_{1\leq j< k}\frac{1}{j^2},\tag{1}$$
that leads to:
$$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=\left(\sum_{k=1}^{+\infty}\frac{H_k}{k^2}\right)\left(\sum_{j=1}^{+\infty}\frac{1}{j^2}\right)-\sum_{k=1}^{+\infty}\frac{1}{k^2}\sum_{1\leq j< k}\frac{H_j}{j^2},\tag{2}$$
Now since:
$$\operatorname{Li}_2(x)+\frac{\log^2(1-x)}{2}=\sum_{k=1}^{+\infty}\frac{H_k}{k}x^k,\tag{3}$$
$$\frac{\log^2(1-x)}{2}=\sum_{k=1}^{+\infty}\frac{H_{k-1}}{k}x^k,\tag{4}$$
follows. By dividing by $x$ and integrating between $0$ and $1$ we get:
$$\sum_{k=1}^{+\infty}\frac{H_{k-1}}{k^2}=\frac{1}{2}\int_{0}^{1}\frac{\log^2(x)}{1-x}dx=\frac{1}{2}\int_{0}^{+\infty}\frac{u^2}{e^u-1}du=\zeta(3),\tag{5}$$
so:
$$\sum_{k=1}^{+\infty}\frac{H_k^{(2)}H_k}{k^2}=2\zeta(2)\zeta(3)-\sum_{k=1}^{+\infty}\frac{1}{k^2}\sum_{1\leq j< k}\frac{H_j}{j^2}.\tag{6}$$
For the last term consider:
$$-\frac{\log(1-xy)}{y(1-xy)}=\sum_{k=1}^{+\infty}H_k x^k y^{k-1}, \tag{7}$$
multiply both terms by $-\log(y)$ and integrate between $0$ and $1$ with respect to $y$:
$$\int_{0}^{1}\frac{\log(y)\log(1-xy)}{y(1-xy)}dy = \sum_{k=1}^{+\infty}\frac{H_k}{k^2}x^k.\tag{8}$$
Multiplying both sides by $-\frac{\log x}{1-x}$ and integrating between $0$ and $1$ with respect to $x$ should do the trick. For the last part it is only required to find an appropriate birational diffeomorphism of the unity square that puts the integral in a nicer form - a sort of "reverse Viola-Rhin method".
|
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|
Book Recommendations for Linear Algebra Proofs I'm taking a graduate Linear Algebra course and have limited experience writing proofs (mostly from a discrete math class). Can anyone recommend good books to teach you how to write proofs for linear algebra? My impression is that they're very different from other kinds of math proofs.
Thanks in advance.
|
Another book you might consider is Curtis' Abstract Linear Algebra. It claims that it could be used for a "first course," but is quite sophisticated (but I have to agree with the claim). It takes some care to introduce the more abstract topics not usually covered in undergrad classes. For instance, exact sequences of maps, defining the determinant via the exterior product, quotient spaces, etc. The majority of the proofs are very easy to follow and the book is a good reference (despite how slim it is).
There's also a book called "The Linear Algebra a Graduate Student Ought to Know." It looks pretty good.
To learn to write proofs, you might look through some of these short "articles" and then do a lot of problems. (I don't recommend a whole book on learning to write proofs). Be sure to have your solutions/proofs critiqued (here for instance!).
|
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For what values of $b\in \mathbb R$ is $\pi-b$ rational? Just a simple short question.
I'm looking for values $b$ such that $\pi-b$ is a rational number.
Obviously $\pi$ is such a number, but are there more?
Edit: $b$ is in $\mathbb R$
|
Clearly, and almost vacuously $b=\pi+q$ for some rational number $q$. You can easily show that these are the only ones.
|
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Finding the area of a region? I sketched out the graph to get this figure but I can't seem to find the area of the shaded region... would one Y = 4 and the other Y = 8?
I, in all honesty, am so flabbergasted with this question, any guidance will be much appreciated. I got 28/3 for the first integral (0 to 4) and 292/3 for the second integral (4 to 16)... I feel like I'm doing this wrong.
Okay, I'm going to be honest, can someone please solve this for me? I got 320/3 as my final answer and yet it's still wrong, I'm just about given up at this point, it's literally my last question of this semester.
|
HINT Find the point of intersection, set the limits $a = 0$ and $b =$ the point you found, set up the integral with those limits and the function $f(x) - g(x)$, and set up another integral from $b$ to $16$ with $g(x) - f(x)$.
|
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Is it true or not : if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic. Is it true or not : if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic.
My try :
$u(z)=u(x,y)$ is harmonic
Define $s=-y$
Let $U := u(\overline{z})=u(x,-y)=u(x,s)$ : $$\frac{\partial U}{\partial x}=\frac{\partial u}{\partial x} \Rightarrow \frac{\partial^2 U}{\partial x^2}=\frac{\partial^2 u}{\partial x^2}$$
And $$\frac{\partial U}{\partial y}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial y} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial y} = - \frac{\partial u}{\partial s} $$
Similarly $$\frac{\partial^2 U}{\partial y^2}= \left[ \frac{\partial }{\partial y} ( \frac{\partial U}{\partial y}) \right] =- \left[ \frac{\partial }{\partial y} ( \frac{\partial u}{\partial s}) \right] = ... = - \left[ - \frac{\partial }{\partial s} ( \frac{\partial u}{\partial s}) \right] = \frac{\partial^2 u}{\partial s^2} $$ Hence
$$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}= 0 \ \ \ (*)$$
Therefore $u(\overline{z})$ is also harmonic.
*My Questions is : Is my try problematic?Does $(*)$ needs justification? *
|
A funciton $\phi$is harmonic if
$$ \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$$
It can be shown that $$ \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} = \frac{\partial}{\partial z}\frac{\partial}{\partial \bar{z}}$$
This means for a function of a complex variable to be harmonic it needs to be an analytic function of just $z$ or an analytic function of just $\bar{z}$. Analytic so that the derivative exists and a function of just one or the other so that the result is zero.
This means that the real part and the imaginary part of said analytic function will each be harmonic functions
The following is true only if $f(z)$ is real when $z$ is real
Note that many cases nothing new is gained in making it a function of $\bar{z}$ since for certain analytic functions this is equivalent to taking the complex conjugate.
$$ f(\bar{z}) = \overline{f(z)} = \overline{u(x,y)+iv(x,y)} = u(x,y)-iv(x,y) $$
So that we don't really get fundamentally new solutions to Laplace's equation.
To obtain the first equality note that when a function is analytic within some neighborhood of a point there is a Taylor series which converges to the function in within some open set containing that point. This allows us to write,
$$f(z) = \sum_{n=0}^N a_n z^n + R_N(z)$$
Where $R_N \rightarrow 0$ as $N \rightarrow \infty$
$$ \overline{ f(z) } = \overline{\sum_{n=0}^N a_n z^n + R_N} = \overline{\sum_{n=0}^N a_n z^n} + \overline{R_N} = \sum_{n=0}^N \overline{a_n} (\overline{z})^n + \overline{R_N}$$
Since $\vert \overline{R_N} \vert = \vert R_N \vert$ we can conclude that $\overline{R_N} \rightarrow 0 $ whenever $R_N$ does. If the coefficientes $a_n$ are real then we can write $\overline{a_n} = a_n$ and conclude that,
$$ \overline{f(z)} = \sum_{n=0}^\infty a_n (\overline{z})^n = f(\bar{z})$$
|
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Sum of difference of numbers in an arrangement of the numbers $0,1,2,\cdots, n$ A seemingly interesting (easy?) problem came to mind and I thought it would be nice to ask your opinion about it.
Suppose we are going to arrange numbers $0$ to $n$ in a row in such a way that the sum of the difference between adjacent numbers is max. How could we accomplish this? How could we determine the maximum sum of the difference? How about the minimum?
As an example let us take the numbers $0,1,2,3,4$. Let us look at some arrangements:
$0,1,2,3,4$; the difference between the adjacent numbers are $1,1,1,1$ so the sum is $4$
$1,3,4,0,2$; the difference between the adjacent numbers are $2,1,4,2$ so the sum is $9$
$1,4,0,3,2$; the difference between the adjacent numbers are $3,4,3,1$ so the sum is $11$
$3,0,4,1,2$; the difference between the adjacent numbers are $3,4,3,1$ so the sum is $11$
I figured that the arrangement that will give us the minimum sum is the arrangement $1,2,3,...,n$ because all the difference would just be $1$ and their sum is $n-1$.
I think the maximum sum can be attained by arranging the numbers in the following way:
Place the number $n$ between $0$ and $1$. Then place $n-1$ next to $0$ and $n-2$ next to $1$ and so on.
An additional question, if we already know the minimum and maximum would it be possible to always get an arrangement that will give a sum for all the values between the minimum and the maximum? I tried it with $0,1,2,3$ and I was able to find arrangements for all the values between $3$ and $7$.
Maybe this problem has been asked before but I can't find it on the net. Any ideas? Thanks!
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I have an idea how to obtain an upper bound which seems to be good. We have to estimate from above the sum $S=\sum_{i=1}^n |a_i-a_{i-1}|$ where the sequence $a_0,\dots, a_n$ is a permutation of the sequence $0,1,\dots,n$. When we open the moduli in the expression for $S$, we obtain $S=\sum_{i=0}^n\varepsilon_i a_i$, where $\sum_{i=0}^n\varepsilon_i=0$ and $\varepsilon_i\in\{-1,1\}$ for $i=0$ or $i=n$ and $\varepsilon_i\in\{-2,0,2\}$ for $1\le i\le n-1$. Suppose that $n=2k+1$ is odd. Now it seems to be true and easily provable that $S$ have the maximum $(n-1)^2/2-1$ when the $\varepsilon$-coefficients at the numbers $0,1,\dots,k-1$ are “-2”, at $k+2,k+3,\dots,n$ are “2”, at $k$ is “-1” and at $k+1$ is “1”. The case when $n$ is even can be considered similarly.
|
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Is $SO_n({\mathbb R})$ a divisible group? The title says it all ... Formally, if
$SO_n(\mathbb R)=\lbrace A\in M_n({\mathbb R}) |AA^{T}=I_n, {\sf det}(A)=1 \rbrace$
and $W\in SO_n(\mathbb R)$, is it true that for every integer $p$, there is a $V\in SO_n(\mathbb R)$ satisfying $V^p=W$ ?
This is obvious when $n=2$, because rotations in the plane are defined
by an angle which can be divided at will.
|
Every compact connected Lie group $G$ (and $SO(n)$ is both compact and connected) is divisible since its exponential map $\exp$ is surjective (see here). By surjectivity of $\exp$, every element $g$ of $G$ has the form $\exp(v)$ (with $v\in {\mathfrak g}$, the Lie algebra of $G$) and, thus, for $h=\exp(v/p)$ we obtain $h^p=g$.
|
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how do you read: "$\lim (n-1)/(n-2) = 1$" The limit of the fraction $(n-1)/(n-2)$ with $n$ approaching $+\infty$ is $1$. But how do you read that, concisely? (say during chalking the formula on a blackboard). Is it acceptable to say "lim", for instance?
I would read "the limit of n minus one over n minus two, with n approaching positive infinity is one".
Would that be correct? But then, how do you deal with the ambiguity regarding parentheses? Would you have to say "the limit of the quantity n minus one over the quantity n minus two.." or do you say "left parenthesis n minus one right parenthesis" ?
|
I tend to read $\lim\limits_{x\to a}f(x)$ as
'The limit as $x$ approaches $a$ of the function $f(x)$ is . . .'
As for dealing with the ambiguity when verbalising a quotient like this, I find that the use of the word 'all' is helpful to distinguish between the possible numerators: $n - 1$ and $1$. I would say
'$n$ minus one all divided by $n$ minus two'
There is still ambiguity as it is not clear whether $n - 1$ is all divided by $n$ or $n - 2$. This is usually avoided by the pacing of the sentence (I'd say '$n$ minus two' faster so that it seems like one object). If I wanted instead to refer to $n - 1$ all divided by $n$ and then minus two from that, I'd change the pacing, putting a little bit of a pause between the final '$n$' and the 'minus two'. You could avoid the ambiguity in this case by instead writing the expression as $-2 + \frac{n-1}{n}$.
|
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Applying Fermat's Little Theorem: $6^{1987}$ divided by $37$ Find the remainder when $6^{1987}$ is divided by $37$.
Because 37 is prime we have: $6^{36}$ mod $37 = 1$. I tried to get a nice combination like: $1987 = 36 * 55 + 7$, so we would have $(6^{36})^{55}6^{7}=6^{1987}$.
Then, I've taken mod $37$, which is: $6^{1987}$ mod $37=1^{55}(6^7$ mod $37)$. I need to find $6^7$ mod $37$. What can I do from here?
Of course, any other method (solution) for finding the remainder would be great.
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Hint $\rm\ mod\ n^2+1\!:\,\ \color{#c00}{n^2}\equiv -1 \ \Rightarrow\, n^{4k+3}\! = n(\color{#c00}{n^2})^{2k+1} \equiv - n.\ $ Yours is special case $\,\rm n=6.$
|
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How can a negative multiplied by a negative give positive? On first look this can seem weird. But I can explain what I am looking for.
We all know from elementary maths that $(-\times-)=+$.
Now, lets say there are 3 cows and I say they will become doubled after one year so $3*2=6$.
And lets say I have $-3$ cows(which is not possible, because I can show $3$ cows but not $-3$) and if I multiply it by $-2$, I get $-3\times -2=6$. How is it possible?, nothing multiplied by nothing equals something?
Some said the reason belongs to philosophy, if yes, what's the idea behind it?
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Multiplication by a positive number is rather intuitive, so $(+ \times +)$ and $(+\times -)$ are relatively easy (for the latter, think you multiply a debt, for example). So the hard one is $(- \times -)$. I may be uneasy to understand this directly, but if you want to keep laws of arithmetic as they are with positive numbers, then
$$a \times (b + c) = a \times b + a \times c$$
With $c=-b$, it yields
$$0 = a \times 0 = a \times b + a \times (-b)$$
$$a \times b = - (a \times (-b))$$
Now, if both $a$ and $b$ are negative, you must have on the left $(- \times -)$, and on the right, the negation of $(- \times +)$, thus a positive number.
|
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Properties of matrix exponential I know that the solution to system $x' = Ax$ is $e^{At}$, and I'm aware of various methods to calculate the exponential numerically. However I wonder if there are some analytical results.
Namely, I'm interested in matrices of type $A_{i,i} = -A_{i+1,i}$; $A_{n,n}=0$. These matrices are the infinitesimal generators of a Markov chain, where you transit from state 1 to state n through various steps. Would it be possible to calculate $[e^{At}]_{n,1}$, i.e. transition probability from state $1$ to state $n$ analytically as a function of time $t$?
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This seems to transform the original problem into a more complicated one... A direct approach is as follows.
For every $1\leqslant i\leqslant n-1$, let $a_i=-A_{i,i}$. The hitting time $T$ of $n$ starting from $1$ is the sum of $n-1$ independent random variables, each exponential with parameter $a_i$. In particular,
$$
E[\mathrm e^{-sT}]=\prod_i\frac{a_i}{a_i+s}.
$$
Assume first that the parameters $a_i$ are distinct. Then the RHS is
$$
\sum_ib_i\frac{a_i}{a_i+s},\qquad b_i=\prod_{j\ne i}\frac{a_j}{a_j-a_i},
$$
hence the distribution of $T$ has density $f$, where, for every $t\geqslant0$,
$$
f(t)=\sum_ib_ia_i\mathrm e^{-a_is}.
$$
In particular,
$$
[\mathrm e^{At}]_{n1}=P[T\leqslant t]=\int_0^tf=\sum_ib_i(1-\mathrm e^{-a_it})=1-\sum_ib_i\mathrm e^{-a_it}.
$$
If some parameters $a_i$ coincide, consider the limit of the expression above when $a_i-a_j\to0$ for some $i\ne j$. This limit is finite and coincides with $[\mathrm e^{At}]_{n1}$. Note that $1-[\mathrm e^{At}]_{n1}$ is always a linear combination of the functions $t\mapsto t^{k-1}\mathrm e^{-at}$, where $a=a_i$ for at least some $1\leqslant i\leqslant n-1$, and $1\leqslant k\leqslant\#\{1\leqslant i\leqslant n-1\mid a_i=a\}$.
|
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find the point which has shortest sum of distance from all points? I want to find a point in the Cartesian plane so that sum of distances from this point to all points in the plane be minimum.
For example we have the points: $(x_1,y_1),(x_2,y_2),(x_3,y_3), . . .(x_n,y_n)$. Now find a point - we call this $(X,Y)$ - so that:
$$\sum_{i=1}^n \sqrt {(x_i−X)^2+(y_i−Y)^2}$$ is minimal.
Thanks in advance.
|
I think what you are looking for is the Geometric median
I'd recommend having a look at this question at stackoverflow
|
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Integration of a rotated triangle: determining the slope of a side I'm working through MIT's OCW single-variable calculus course. The problem is taken from Problem set 7: Question 4B-5.1
Q: Find the volume of an equilateral triangle with side length a which is rotated around one of its sides.
I am actually doing the calculus side okay, I believe. My difficulty is with the trigonometry/geometry setting up the problem. Here is how I set up the integral:
$$
2 \int_0^\frac{a}{2} \pi(\frac{\sqrt{3}}{a}x)^2dx
$$
My integrand is apparently wrong. I've calculated $y$ to be a function of $x$ such that $y = \frac{\sqrt{3}}{a}x$. The solution sets it up the integral exactly the same, except that $y = \sqrt{3}x$.
Here is how I derived the integrand:
Let the equilateral triangle sit with one side along the x-axis and its leftmost vertex at (0,0) and its rightmost vertex at (a, 0). The strategy will be to split the triangle in half, integrate the rotated solid around the x-axis, and multiply by two to get the whole triangle.
So the function to be integrated is a simple linear f(x) from 0 to $\frac{a}{2}$. Great. But what is $f(x)$? Here's where I go wrong.
I derived the slope of the line of f(x) by observing that the vertex at (0,0) is $\frac{\pi}{3}$ radians. Using sohcahtoa, I calculated the height of the triangle (i.e., $f(\frac{a}{2})$) as:$$\sin(\frac{\pi}{3})$$
To then derive the slope of f, I simply did rise-over-run, i.e. (plugging in for $\sin\frac{\pi}{3})$:
$$\frac{\sqrt{3}}{2}\div\frac{a}{2} = \frac{\sqrt{3}}{a}$$
Which is the slope of the function f(x). I will leave it here, as I don't think the calculus is the problem. Clearly, I've done something wrong in the way I derived the slope of $f(x)$. What am I missing? What would have been the correct way to calculate $f(x)$ to arrive at $y = \sqrt{3}x$?
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You were close, but height needs to be $a\sin\left(\frac \pi 3\right)$. Then slope is given by $\dfrac {\frac {a\sqrt 3}2}{\frac a2} = \sqrt 3$.
|
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Finding the value of a logarithmic expression involving an infinite GP Find the value of $(0.16)^{\displaystyle\log_{2.5}(\frac13+\frac1{3^2}+\frac1{3^3}+\cdots)}$.
I could solve the series. It gave $$(0.16)^{\log_{2.5}0.5}$$
Unable to proceed from here.
|
HINT:
$$0.16=\frac4{25}=\left(2.5\right)^{-2}$$
$$\text{Now, }\displaystyle a^{\log_am }=m$$
|
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How to prove that triangle inscribed in another triangle (were both have one shared side) have lower perimeter? This question looks really simple, but to my (and my co-workers) frustration we were not able to prove this in any way. I tried all triangle formulas known to me but I feel I'm missing the point, and proof will be or much simpler or much more complicated than what I tried.
So, the question:
Given a triangle ABC and point P inside that triangle, prove that for triangle APB the following inequality holds:
|AB| + |BC| > |AP| + |PC|
(Actually it doesn't matter for me if it's > or >=).
|
There is really an elementary proof of a more general result: if one polygon fits within the other (boundaries may touch) then its perimeter is smaller than that of the bigger polygon
http://www.cut-the-knot.org/m/Geometry/PerimetersOfTwoConvexPolygons.shtml
The idea is that the sides of the small polygon may be extended, trimming the bigger polygon until the sides of one are on the sides of the other so that side-by-side inequalities could be summed up to get an inequality for the perimeters
|
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Hunt for a function. I am looking for any nontrivial function $f(z): \mathbb{C}\to\mathbb{C}$ such that:
*
*$f(z)$ is an entire function.
*A $z_p\in\mathbb{C}$ exists for which $\Re(f(z))\geq\Re(f(z_p))~\forall z\in\mathbb{C}$ with $\infty>\Re(f(z_p))>-\infty$.
If you do not have an example, but can give me some strategy on how to look for such a function, please share your thoughts. Thanks!
EDIT:
Ok, turns out the condition entire is too restrictive. Any idea by how much I should relax the restrictions to get the second property (and preferably the same for the imaginary part at the same $z_p$)?
|
Such a function must be constant.
Consider $g(z) = \exp(-f(z))$. We have
$$
\left|g(z)\right| = \exp(-\Re(f(z)) \le \exp(-\Re(f(z_p)).
$$
Since $g$ is entire and bounded, it must be constant. Hence $f$ is constant too.
|
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Limit with missing variables
Find the values for $a$ and $b$ such that
$$\lim_{x \to 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = 3$$
Basically what I did so far was I started by multiplying by the conjugate. and obtained $$\frac{a+bx-3}{x(\sqrt{a+bx}+\sqrt 3)}$$ I don't know what to do after this.
|
Being finite, numerator $\to 0\,\Rightarrow\,a = 3,$ so it is $\,\displaystyle\lim_{x\to0}\dfrac{f(x)-f(0)}x = f'(0),\ \ \ f(x) = \sqrt{3+bx}$
|
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Question about differentiation of series Suppose $$f(x)=\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x}{n}), \quad x\in[0,\infty).$$ I need to show that $f$ is differentiable on $(0,\infty).$
proof: I try to show differentiability using the classical defintion of limit.
$$\lim _{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ $$=\lim _{h\rightarrow 0} \bigg(\frac{\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x+h}{n})-\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x}{n})}{h}\bigg)$$ $$ =\lim _{h\rightarrow 0} \sum_{n=1}^\infty (-1)^{n+1} \frac{\bigg(\ln (1+\frac{x+h}{n}) -\ln (1+\frac{x}{n})\bigg)}{h}$$ $$=\sum_{n=1}^\infty (-1)^{n+1} \lim_{h\rightarrow 0} \bigg(\frac{ \ln (\frac{n+x+h}{n}) -\ln (\frac{n+x}{n})}{h}\bigg)$$ $$=\sum_{n=1}^\infty (-1)^{n+1} \lim_{h\rightarrow 0} \bigg(\frac{1}{\frac{n+x+h}{n}.\frac{1}{n}}-\frac{1}{\frac{n+x}{n}.\frac{1}{n}}\bigg)\mbox{, By L'Hospital's rule.}$$ $$=\sum_{n=1}^\infty (-1)^{n+1} .0 =0 $$
Thus, limit exists. Since $x$ was arbitrary, the claim follows.
Only problem I have is in justifying the exchange of limit and sum.
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$$ \lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\left(\ln(\frac {n+x+h}n) - \ln(\frac {n+x}n)\right)}{h} \\
= \lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\ln(\frac {n+x+h}{n+x})}{h} \\
= \lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\ln(1+\frac {h}{n+x})}{h} $$
Now use the formula $\ln(1+a) = a + O(a^2)$ as $a\to 0$. You then will get a term where the $h$'s cancel, plus a term that involves an absolutely converging series times $h$.
|
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Finding the derivative using quotient rule... $$\frac{\text{d}}{\text{dt}}\dfrac{2(t+2)^2}{(t-2)^2}$$
I applied the quotient rule:
$$\dfrac{[2(t+2)^2]'(t-2)^2-2(t+2)^2[(t-2)^2]'}{(t-2)^4}$$
$$\dfrac{4(t+2)(t-2)^2-2(t+2)^22(t-2)}{(t-2)^4}$$
$$\dfrac{4(t+2)(t-2)-4(t+2)^2}{(t-2)^3}$$
This was part of a problem where I needed to find the second derivative of a parametric curve but I am stuck on finding this derivative. I typed this problem into wolfram alpha and it gave me $\dfrac{-(16 (x+2))}{(x-2)^3}$, I've been working on this problem for the past hour and can't figure out what I am doing wrong, could someone please explain how to do this?
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Your answer is correct, Wolfram is just simplified.
$$ \frac{(t-2)^2 \cdot 4(t+2) - 2(t+2)^2 \cdot 2(t-2)}{(t-2)^4}$$
$$=\frac{(t-2)(t+2) \cdot 4(t-2) - 4(t+2)}{(t-2)^4}$$
$$= \frac{(t+2) \cdot (4t - 8 - 4t - 8)}{(t-2)^3}$$
$$= \frac{-16(t+2)}{(t-2)^3}$$
|
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Why $\int{\log^2(2\sin(\pi x))}dx\neq\frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$? Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:
$$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$
UPDATE:
Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:
$$\int{\log^2(2\sin(\pi x))}dx = \frac{\log^3(2\sin(\pi x))\sin(\pi x)}{3\pi \cos(\pi x)}$$
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Using Chain rule:$$\frac{d }{dx}\log^2(2\sin(\pi x))=2\log(2\sin(\pi x))\times[2\times\pi\times\cos(\pi x)]\times\frac{1}{2\sin(\pi x)}$$
|
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Example of a non locally-compact space with a dense locally compact subspace. Let $X$ be a topological space such that $A \subset X$ is a dense subspace which is locally compact and $B \subset X$ is a dense subspace which is not locally compact (at all of its points).
Is it possible to find such $X$? if it is, an example?
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1)If $X$ must be Hausdorff, take $X=\mathbb{R}$, $A=X-\{0\}$, $B=\mathbb{Q}$.
2)If $X$ must be a non-locally compact Hausdorff space, let $Y=\{1/n|n\in \mathbb{N}^*\}$ and $X=(\{0\}\times \mathbb{Q}) \cup (Y\times \mathbb{R})$, $X$ is not locally compact at the point $(0,0)$.
Let $A=Y\times \mathbb{R}$, $A$ is dense in $X$ and is locally compact.
Let $B=Y \times \mathbb{Q}$, $B$ is dense in $X$ and $B$ is not locally compact at all of its points.
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Proof that $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}$ Being $\mathbb G_n$ the roots of unity for $n \in \mathbb N$, prove that $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}.$
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Hint
Recall
$$z\in\mathbb G_n\iff z^n=1$$
By double inclusion
*
*if $d|n$ prove that
$$z\in\mathbb G_d\Rightarrow z\in\mathbb G_n$$
and hence which inclusion we can deduce?
*Use the Bézout's theorem:
$$d=\gcd(m,n)\iff \exists u,v\in\mathbb Z,\; um+vn=d$$
to get the other inclusion.
|
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find a the following series this is just somthing i thought about (i dont know if there is an answer)
Let $ \sum_{n=1}^{\infty}a_n \to L_1$
($a_n $ is a positive sequence)
find a sequence $b_n$ such that:
$\lim_{n\to\infty} \dfrac{b_n}{a_n} = \infty$
and $\sum_{n=0}^{\infty}b_n = L_2$ (the series converges)
if there is no solution can this be proven?
if we add the fact that $a_n$ is monotone decending would that help?
thanks alot!
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NOTICE
This is an answer to the original question. While I was writing the answer the OP changed the condition $a_n/b_n\to\infty$ to $b_n/a_n\to\infty$.
Take $b_n=a_n^2$. Since $\sum a_n$ converges, $a_n\to0$, and $a_n/b_n=1/a_n\to\infty$. Also, $a_n$ is bounded. Let $A$ be an upper bound. Then $0\le b_b\le A\,a_n$, so that $\sum b_n$ converges.
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Show that $A= ([0,\sqrt2] \cap \Bbb Q ) \subset \Bbb Q $ is not compact. We have $\Bbb Q$ equipped with the Euclidean Metric.
Show that $A=([0,\sqrt2] \cap \Bbb Q ) \subset \Bbb Q $ is not compact.
How would you go about showing this?
You can make on open cover $\{O_n\}= ((-\infty, \sqrt2 -\frac1n) \cap \Bbb Q)$ for $n\ge1$
$ A \subset {O_n}$
How would you show there is no finite subcover?
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If a subset $A$ of $\mathbb R$ is compact, then it is also closed. In particular, if a ssequence $\{x_n\}_{n\in}\subset A$ converges to $x$, then $x\in A$. Let
\begin{align}
x_1=&1\\
x_2=&1.4 \\
x_3=&1.41 \\
x_4=&1.414 \\
x_5=&1.4142 \\
etc.
\end{align}
I.e.,
$$
x_n=\frac{\lfloor10^n\sqrt{2}\rfloor}{10^n},
$$
Clearly $x_n\to\sqrt{2}$, as $|x_n-\sqrt{2}|<10^{-n}$, and $\sqrt{2}\not\in A$. Hence $A$ is not compact.
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Inductive proof of $\,9 \mathrel| 4^n+6n-1\,$ for all $\,n\in\Bbb N$ Prove that for all $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$.
I have shown the base case if $n=0$ (which is a natural number for my course).
I'm now trying to prove that $9k=4^n+6n-1$.
I substituted $n+1$ for $n$, and have $4^{(n+1)}+6(n+1)-1$, which equals $4*4^n+6n+5$.
I'm stuck. Is that not the right way to go? I don't know how to get from the “$+5$” to “$-1$” so I can substitute $9k$ to the right side of the equation. Any help would be greatly appreciated.
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It's really simple this is how I'd approach it:
$4^{n+1}+6(n+1)-1=4(4^n+6n-1)-18n+9=4(4^n+6n-1)-9(2n-1)$
and since you've already proved that $4^n+6n-1=9k$ then you will have: $4(4^n+6n-1)-9(2n-1)=9k*4-9(2n-1)$
$=9(4k-(2n-1))$ and since $k$ and $2n-1$ are both integers then you can let $k'=4k-(2n-1)$ .
Now you'll have that $4^{n+1}+6(n+1)-1=9kk'$ which is a multiple of 9. So you can conclude for all natural numbers $n$ that $4^n+6n-1$ is divisible by $9$.
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Is a set that is an abelian group under addition and a group under multipliation a field? I suspect the answer to my question is yes, but I'm just checking my understanding. If we have a set which is an Abelian group under addition and a group under multiplication is it then defined as a field?
Cheers Matt
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In general, no, because the distributive laws may not hold.
With that and a minor modification, yes. The multiplicative group has to leave out $0$, because $0$ won't be invertible.
So if you have a ring (with identity) such that the nonzero elements are an abelian group, then yes, you have a field.
If the nonzero elements are just a group then you have a division ring.
|
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|
positive harmonic function has a zero limit at a point on the boundary Let $u$ be a positive harmonic function in $\{ \Re{z} > 0\}$ such that
$\lim_{r \rightarrow 0^+} u(r) = 0 $.
Prove that then $\lim_{r \rightarrow 0^+} u(re^{i\theta}) = 0 $ for any $\theta \in (-\pi/2, \pi/2)$.
What I have tried is to consider the equivalent problem by transforming the domain to the unit disc using the Moebius map $z \rightarrow \frac{z-1}{z+1}$. Then we have
$\lim_{r \rightarrow -1^+} u(r) = 0 $ and we need to prove that this limit is zero when approaching along the arcs of other circles which intersect the unit circle perpendicularly at $-1$ or in other words the limit is the same when approaching from different non-tangential directions. Note also we have a simply connected domain and hence a holomorphic function, $f$ exists whose real part is $u$. Also since u was positive, hence $f$ maps the unit disc to itself which means it is bounded. This in turn means that $f$ has a finite limit a.e. on the boundary. Any ideas how to proceed or how to prove this otherwise?
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The assumption that $u$ is positive should bring to mind Harnack's inequality. Let's state its special case for the disk: if $u$ is positive and harmonic in $\{z:|z-z_0|<R\}$, then
$$\frac{R-|z-z_0|}{R+|z-z_0|}u(z_0)\le u(z) \le \frac{R+|z-z_0|}{R-|z-z_0|}u(z_0)$$
for all $z$ in the disk.
The precise form of constants does not matter; the fact that they depend only on $|z-z_0|/R$ is a consequence of the scale invariance of the Laplace equation.
Apply the above with $z=re^{i\theta}$ and $z_0=R=r\sec\theta$. Observe that $|z-z_0|/R=|\sin\theta|<1$, independently of $r$. Thus,
$$u(re^{i\theta})\le \frac{1+|\sin\theta|}{1-|\sin\theta|}u(r\sec\theta)$$
from where the conclusion $u(re^{i\theta})\to 0$ follows.
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What is the big picture behind AKS algorithm? Despite a number of question on AKS algorithm here, there does not seems to anything related to the idea behind it (for those who don't know, AKS primality testing is found in PRIMES is in P).
I read through the paper, check all the step for correctness (and also fix a few small errors). Yet I still can't understand it at all, in the sense that I still can't see the big picture here. From my perspective, it looks like a bunch of unintuitive flashes of insight that just get thrown together until they work.
So can someone explain the big idea behind it please?
I would appreciate if someone can explain it fully, but here are a few specific questions to get you started:
-How does the concept of "introspective" come up?
-Why is it make sense to look at the group $G$ and $\mathcal{G}$?
-How much "wiggle room" is there for the bound $o_{r}(n)>(\log_{2}n)^{2}$ (specifically the RHS)?
-How could one have come up with the proof of an upper bound for $|\mathcal{G}|$?
-How much "wiggle room" is there for $\lfloor\sqrt{\phi(r)}\log_{2}n\rfloor$?
Thank you for your help.
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There is an expository article by Granville titled"It is easy to determine whether a given integer is prime," which answers exactly the question you are asking. The article is worth the read, and it won the 2008 Chauvenet prize for its exposition.
Edit: This article was also referenced by Will Jagy in the comments.
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Simple functional equation: $\frac 1 2 [\alpha(x - 1) + \alpha(x + 1)]$, $\alpha(0) = 1$, $\alpha(m) = 0$ I have a simple functional equation:
$$
\alpha(x)
=
\frac 1 2 [\alpha(x - 1) + \alpha(x + 1)]\,,
\qquad \alpha(0) = 1\,,\quad\alpha(m) = 0
$$
I know it has a linear solution $\alpha(x) = ax + b$ but I don't have any idea how to prove that this is the only solution. Is there any way I can derive this solution from the initial equation?
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In your last comment to date, you rightly observe that for $b(x)=a(x)-a(x-1)$ you get the equation
$$b(x+1)=b(x)$$
Its solution are all functions with period 1.
Now
$$a(x+n)=a(x+n-1)+b(x)=a(x+n-2)+2\,b(x)=...=a(x)+nb(x)$$
using the periodicity of $b$. This leads to the idea to consider
$$c(x)=a(x)-x\,b(x).$$
It satisfies the discrete dynamic
\begin{align}
c(x+1)&=a(x+1)-(x+1)\,b(x+1)=a(x)+b(x+1)-(x+1)\,b(x+1)\\
&=a(x)-x\,b(x)=c(x),
\end{align}
so it is again a periodic function with period $1$. The general solution has thus the form
$$a(x)=c(x)+x\,b(x)$$
with $b$ and $c$ any 1-periodic functions. This specializes to the linear solution in the case that $b$ and $c$ are constant functions.
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Dirichlet series for inverse of Eta function We know that $$ \frac{1}{\zeta (s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} $$
but what happens with $$ \frac{1}{\eta (s)} = \sum_{n=1}^{\infty} \frac{b(n)}{n^{s}} $$
with $ \eta (s) = (1-2^{1-s})\zeta (s) ?$
Can I evaluate the coefficients $ b(n) $ ? Perhaps I should apply the mobius transform to the sequence $ a(n)= (-1)^{n} .$
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$$\frac{1}{\eta(s)}=\frac1{(1-2^{1-s})\cdot\zeta(s)}=\frac1{1-2^{1-s}}\cdot\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}=\sum_{n=1}^{\infty}\frac{b(n)}{n^{s}}\iff b(n)=\frac{\mu(n)}{1-2^{1-s}}$$
|
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$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can I approach
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Apart from the Prosthaphaeresis Formulas already mentioned with the unmentioned assumption that $\displaystyle \sin\frac{\alpha+\beta}2\ne0$
we can try as follows :
Rearranging we have $\displaystyle\sin\alpha+\sqrt3\cos\alpha=\sqrt3\cos\beta-\sin\beta$
As $\displaystyle 1^2+(\sqrt3)^2=4,$ we write $1=2\sin30^\circ,\sqrt3=2\cos30^\circ$ to get
$\displaystyle2\sin30^\circ\sin\alpha+2\cos30^\circ\cos\alpha=2\cos30^\circ\cos\beta-2\sin30^\circ\sin\beta$
$\displaystyle\implies \cos(\alpha-30^\circ)=\cos(30^\circ+\beta)$
$\displaystyle\implies \alpha-30^\circ=n360^\circ\pm(30^\circ+\beta)$
Taking the '-' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ-(30^\circ-\beta)\implies \alpha=n360^\circ-\beta$
This makes both sides equal for all $\alpha,$ so no solution available from here
Taking the '+' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ+(30^\circ-\beta)\implies \alpha-\beta=n360^\circ+60^\circ$
Now, use $\displaystyle2\cos^2A=1+\cos2A$ as $\displaystyle\cos2A=2\cos^2A-1$ for $\displaystyle A=\frac{\alpha-\beta}2$
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Find $\displaystyle \lim_{x \to 0} \frac{x \cdot \operatorname{cosec}(2x)}{\cos(5x)}$ I am having difficulties to find the limit for
$$\lim_{x \to 0} \frac{x \cdot \operatorname{cosec}(2x)}{\cos(5x)}$$
I tried to get rid of $ \operatorname{cosec} $ fist
$$\lim_{x \to 0} \frac{\dfrac{x}{\sin(2x)}}{\cos(5x)}$$
Probably I should get it to a point where I could make use of $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$ but I don't know how to continue.
Maybe if one could give me a hint for the next step?
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HINT : You can use $\lim_{x\to 0}\frac{\sin(2x)}{2x}=1.$
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|
Solving special boolean equation set I have boolean equation sets that look like this (where ^ means xor):
eq 1: x1^x3^x5^x6^x9^x10^x11^x13^x17^x18 = 0
eq 2: 1^x1^x3^x10^x12^x17 = 0
eq 3: 1^x2^x3^x5^x8^x10^x14^x16 = 0
eq 4: 1^x4^x5^x6^x7^x8^x10^x16^x17 = 0
eq 5: x2^x5^x8^x11^x13^x14^x17^x18 = 0
(This example has 18 vars and 5 equations, but imagine a thousand variables with a hundred equations).
I can easily find solutions for this by successive subtitutions, but is there any way to quickly generate solution(s) with the least number of variables set to true ?
Thanks!
|
We can phrase the problem as follows:
Given $M \in \mathbb{F}_2^{m\times n}$ and $b \in \mathbb{F}_2^m$ with $n>m$, find the solution $x \in \mathbb{F}_2^n$ to $Mx=b$ for which $\sum_{i=1}^n x_i$ is minimized.
That is, this may be regarded as a "linear programming" problem, of sorts. Below is my initial attempt at a solution, which was not particularly fruitful.
We can write this particular system as the matrix equality
$$
M\,x = b
$$
Where $M$ is the matrix of coefficients (either $1$ or $0$) of $x_1,\dots,x_{18}$, $x = \pmatrix{x_1&\dots&x_{18}}^T$, and $b = \pmatrix{0&1&1&1&0}^T$ (corresponding to whether the equation contains a "1^"). With row reduction, we can determine how many of $x_1,\dots,x_{18}$ are "free variables", i.e. variables that may be arbitrarily set to true or false. The key here is to note that we define multiplication and addition over these elements/coefficients in $\{0,1\}$ as follows:
$$
a+b = a\text{ XOR } b\\
a\cdot b = a \text{ AND } b
$$
This is the definition of $\mathbb{F}_2$.
Let's take an example. Suppose we have the system
x1 ^ x2 ^ x3 == 0
x2 ^ x3 ^ x4 == 1
x3 ^ x4 ^ x5 == 1
We can write this as
$$
\pmatrix{
1&1&1&0&0\\
0&1&1&1&0\\
0&0&1&1&1}
\pmatrix{x_1 \\ \vdots \\ x_5} =
\pmatrix{0 \\ 1 \\ 1}
$$
We can now row reduce the augmented matrix $(M\mid b)$ to get
$$
\pmatrix{
1&1&1&0&0&0\\
0&1&1&1&0&1\\
0&0&1&1&1&1}
\implies\\
\left(
\begin{array}{cccccc}
1 & 0 & 0 & -1 & 0 & -1 \\
0 & 1 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 \\
\end{array}
\right)=
\left(
\begin{array}{cccccc}
1 & 0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 \\
\end{array}
\right)
$$
That is, the system may be rewritten as
x1 ^ x4 == 1
x2 ^ x5 == 0
x3 ^ x4 ^ x5 == 1
So that $x_4,x_5$ may be freely chosen, and the values of $x_1,x_2,x_3$ may be solved for as
x1 == 1 ^ x4
x2 == x5
x3 == 1^x4^x5
Hope that makes sense.
NOTE: this may not be as useful for the particular problem as I originally thought.
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How do you prove that $\Bbb{Z}_p$ is an integral domain? Let $\Bbb{Z}_p$ be the $p$-adic integers given by formal series $\sum_{i\geq 0} a_i p^i$. I'm having trouble proving that it's an integral domain.
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As $\Bbb Z_p/p^n\Bbb Z_p\cong\Bbb Z/p^n\Bbb Z$, to show $a,b\ne0\Rightarrow ab\ne0$ it suffices to show that $ab\not\equiv0$ mod $p^n$ for a high enough power $p^n$ given $a,b\ne0$ (see vadim's hint in the comments).
|
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|
Queueing model - expected outflow Can anybody please help me how to tackle this question?
We have one server.
The service time is random with mean 1 minute
The arrival rate is constant with 3 customers/minute, but they leave if the server is occupied.
A)
Assume the server is empty. How long would it take before one enters the system?
B)
What is the expected number of customers served after 1 hour?
I have a hard time of figuring out how to estimate the fraction of time the server is empty; and obviously then there is not served 1 customer per minute on average.
|
The arrival rate is $3$ per minute, so $1$ every $20$ seconds. That means you have to wait on average $10$ seconds for the next customer. Each customer takes $60$ seconds of time, so in total, thats $1$ customer every $70$ seconds. Per hour, you get $\frac 1{70}\cdot3600=\frac{360}7$.
Not that I am assuming that arrival rates are constant, but because serving time is random, two consecutive waiting times are not correlated.
|
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|
Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} \right) > \dfrac{\sqrt{2}}{2}$
To find bounds: $\sin(x) = \dfrac{\sqrt{2}}{2},\ x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$
Resolving to $x + \dfrac{\pi}{3} \implies \dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$
But wolfram gives a way different result, where's my mistake ?
|
The inequality
$$
\sin \left( x - \frac{\pi}{3} \right) > \frac{\sqrt{2}}{2}
$$
implies that
$$
\frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}.
$$
Adding $\frac{\pi}{3}$ to all three expressions yields
$$
\frac{7\pi}{12} < x < \frac{13\pi}{12}.
$$
If you impose the initial restriction, then the upper bound is $\pi$.
|
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|
Is the power of 1/2 same thing as principal square root? $\sqrt{9} = 3$
9 has 2 square roots: 3 and -3.
What is $9^\frac12$? Is $9^\frac12 = \sqrt{9} = 3$ or is $9^\frac12 = \pm3$?
|
$a^b$ is always positive for $a\in\Bbb R^+$ and $b\in\Bbb R$. This is because $f(x)=a^x$ satisfies the functional equation $f(x+y)=f(x)f(y)$ (i.e. $a^{x+y}=a^xa^y$). It follows that $f(2x)=f(x)^2,$ or $$a^x=f(x)=f\left(\frac x2\right)^2\ge 0$$
Consequently, $9^\frac12\ge 0\implies 9^\frac12=3$
|
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Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\
\\
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\
\\
=& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\
\\
=& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}}
\end{align*}$$
Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?
|
$$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}=\lim_{x \to \infty} \frac{x^{1/3}-x^{1/5}}{x^{1/3} + x^{1/5}}=$$
$$=\lim_{x \to \infty} \frac{1-x^{1/5-1/3}}{1 + x^{1/5-1/3}}=\lim_{x \to \infty} \frac{1-x^{-2/15}}{1 + x^{-2/15}}=\lim_{x \to \infty} \frac{1-\frac{1}{x^{2/15}}}{1 + \frac{1}{x^{2/15}}}=1$$
|
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|
How to find the graph of $e^{2x}$ and $\ln y$ I'd like to know how find the graph I just only know how to draw the common function graph as circle or ellipse but I don't know how to draw.
For example
$e^{2x}$ and $\ln y$.
Note:the interval for $e^{2x}, \;[0,\ln 2].$
How can I determine $\ln 2$ on the graph?
It's better if somebody help me to find resource to learn more about graph -online please
Thanks all
|
You should know the graph of $y=e^x$.
Then, the graph of $y=e^{2x}$ is the 'shorten' graph of $y=e^x$ in the direction of the $x$ axis . You'll see the both graphs here.
Also, note that $e^0=1, e^{2\ln 2}=4$, which means the graph of $y=e^{2x}$ passes the two points $(0,1),(\ln 2, 4).$ However, we don't know the exact value of $\ln 2\approx0.693.$ see here.
|
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|
Diagonalization and eigenvalues Let $A$ $\in M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases when
$A$ is diagonalizable over $\mathbb R$:
a. when $A^2 = A$;
b. when $(A - 3I)^2 = 0$;
c. when $A^2 + I = 0$.
I could eliminate c. by using the equation $\lambda^2+1$ and showing if a matrix has to satisfy this then it has to be diagonal.
I am also in doubt whether all the eigen values of $A$ satisfy this or not.
|
Not sure what the question is, so I will take a stab at the answer. I am assuming you want to know what additional condition is needed to ensure that it is diagonalizable.
a) $A^2=A$
In this case the eigenvalues of $A$ have to be zero or one. Since we need three eigenvalues,
at least one of them is repeated. Now Jordon block of the repeated eigenvalue has to be diagonal for (a) to hold. Hence $A$ is diagonalizable.
b) $(A-3I)^2=0$
This need not be diagonalizable, unless $A-3I=0$. If this is not true then A has to be similar to
$$A = \begin{pmatrix}3 & 1 & 0\cr 0 & 3 & 0\cr 0 & 0 & 3\end{pmatrix}$$
which can't be diagonalized.
c) A^2+I = 0
This can't be diagonalized in the field of real numbers. This is because the eigenvalues of any diagonal matrix are the values in the diagonals. However, $A$ has eigenvalues $\pm \sqrt{-1}$.
|
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|
What values can $v-e+f$ attain if $G$ is a planar (non connected) graph? Let $G=(V,E)$ be a planar graph and choose planar representation.
If $G$ is connected, then according to Euler's formula, we have $$v − e + f = 2,$$ were $v$ is the number of vertices, $e$ the number of edges and $f$ the number of regions bounded by edges, including the outer, infinitely large region.
I was wondering which values $v-e+f$ can assume if $G$ is planar, but not connected. After some drawing, I feel that it can be any integer larges than 1, but I cannot prove it.
Is there someone who can help me out?
|
note that one of the faces of a planar graph is the unbounded region "outside" the graph. this is in common to the separate connected components, so you can then sum over components and obtain:
$$ v-e+f = 1 + C
$$
where $C$ is the number of connected components.
|
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|
Map $\{x+iy \mid x^2+y^2<1 \text{ and } x^2 + (y-1)^2<2\}$ conformally to UHP From an old qualifying exam:
Let $D$ be the domain $$D :=\{x+iy \mid x^2+y^2<1 \text{ and } x^2 +
(y-1)^2<2\}.$$
*
*Map the domain onto the upper half-plane.
*Obtain a function $f(z)$ analytic in the domain $D' := D \cap \{x+iy \mid x>0\}$ and which takes on the boundary values
$\text{Re}f(z) = -1$ on the segment of the imaginary axis $-1<y<1$,
and $2$ on the bounding circular arcs in $D'$, excluding the points
$z=i$ and $z=i-\sqrt{2}i$. Further $\text{Im}f(0)=0$. Is $f(z)$
unique?
(For the first, I have to assume they want a conformal mapping, but they did not indicate it...go figure. For the second, one of the arcs is not "circular", but I guess I know what they mean.)
Let's focus on the first part for now; maybe if I get that I can get the second part. I have struggled to come up with some ideas for the first one. I don't know of any results about conformal mapping and parabolae. Maybe I should look at the polar form of a parabola?
|
First from the definition it seems $D := \{|x^2+y^2|<1\}$ or the unit disc itself. If that is not a mistake, then the map $z \rightarrow i \dfrac{1+z}{1-z}$ sends the unit disc to the upper half plane conformally.
However, a more interesting problem would be to map $D:= \{|x^2+y^2| >1 \text{ and } x^2 + (y-1)^2 < 2\}$ to the UHP. Here D would be the simply-connected region bounded between the two circles that are tangent to each other at $z = -i$. Since this is a special point, it would help to map this point to infinity using $z \rightarrow \dfrac{1}{z+i}$. Then the unit circle gets mapped to the line $z = - i/2$ and the larger circle gets mapped to the line $z = -i/4$. So we get an infinite strip parallel to the real axis. This is because due to conformality the circles which were tangent to each other at $-i$ get mapped to generalized circles (in this case straight lines) which are tangent to each other at infinity, which means two parallel straight lines. From here the steps are easy.
*
*Translate the strip vertically up using $z \rightarrow z+ i/2$.
*Dilate the strip using $z \rightarrow 4\pi z$.
*Finally use $z \rightarrow e^z$ to send the above domain to the UHP.
|
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|
Who came up with the $\varepsilon$-$\delta$ definitions and the axioms in Real Analysis? I've seen a lot of definitions of notions like boundary points, accumulation points, continuity, etc, and axioms for the set of the real numbers. But I have a hard time accepting these as "true" definitions or acceptable axioms and because of this it's awfully hard to believe that I can "prove" anything from them. It feels like I can create a close approximation to things found in calculus, but it feels like I'm constructing a forgery rather than proving.
What I'm looking for is a way to discover these things on my own rather than have someone tell them to me. For instance, if I want to derive the area of a circle and I know the definition of $\pi$ and an integral, I can figure it out.
|
Contrary to a common misconception, one will not find an epsilon, delta definition of continuity in Cauchy even if you look with a microscope. On the other hand, you will find his definition of continuity in terms of infinitesimals: every infinitesimal increment $\alpha$ necessarily produces an infinitesimal change $f(x+\alpha)-f(x)$ in the function. More specifically, the recent translation
Bradley, Robert E.; Sandifer, C. Edward Cauchy's Cours d'analyse. An annotated translation. Sources and Studies in the History of Mathematics and Physical Sciences. Springer, New York, 2009
contains the following material on Cauchy's definition. Cauchy's Section 2.2 is entitled Continuity of functions. Cauchy writes: "If, beginning with a value of $x$ contained between these limits, we add to the variable $x$ an infinitely small increment $\alpha$, the function itself is incremented by the difference $f(x+\alpha)-f(x)$",
and states that "the function $f(x)$ is a continuous function of $x$ between the assigned limits if, for each value of $x$ between these limits, the numerical value of the difference $f(x+\alpha)-f(x)$ decreases indefinitely with the numerical value of $\alpha$." Cauchy goes on to provide an italicized definition of continuity in the following terms:
The function $f(x)$ is continuous with respect to $x$ between the given limits if, between these limits, an infinitely small increment in the variable always produces an infinitely small increment in the function itself.
|
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|
Sequential sums $1+2+\cdots+N$ that are squares While playing with sums $S_n = 1+\cdots+n$ of integers,
I have just come across some "mathematical magic"
I have no explanation and no proof for.
Maybe you can give me some comments on this:
I had the computer calculating which Sn are squares,
and it came up with the following list:
Table
row $N$ sum($1+\cdots+N$) M (square root of sum)
r=1 N=1 sum=1 M=1
r=2 N=8 sum=36 M=6
r=3 N=49 sum=1225 M=35
r=4 N=288 sum=41616 M=204
r=5 N=1681 sum=1413721 M=1189
r=6 N=9800 sum=48024900 M=6930
Of course we have $1+\cdots+N = \frac{N(N+1)}{2}$,
but this gives no indication for which N the sum $1+\cdots+N$ is a square.
Can you guess how in this table we can calculate the entries in row 2 from the entries in row 1?
Or the entries in row 3 from the entries in row 2?
Or the entries in row 4 from the entries in row 3?
Or the entries in row 5 from the entries in row 4?
I looked at the above table and made some strange observations:
*
*The value of the next M can be easily calculated from the previous entries:
Take the M from the previous row, multiply by 6 and subtract the M from two rows higher up.
$M(r) = 6*M(r-1)–M(r-2)$
How is this possible?
The S(r) we calculate as $S(r) = M(r)^2$. Note that we do not know whether this newly constructed
number $S_r$ is in fact of the type $1+\cdots+k$ for some $k$.
*The value of the next N can be calculated as
N(r) = Floor($M(r)*\sqrt 2$),
where Floor means “rounding down to the next lower integer“.
Somewhat surprising, $S(r)$ is the sum $1+\cdots+N(r)$ !
*It looks as if outside the entries in the above table there are no other cases.
With other words, the method $M(r) = 6*M(r-1)–M(r-2)$
seems to generate ALL solutions n where the sum $1+\cdots+n$ is a square.
Problems:
Is there a proof for any of the three observations?
Do observations 1 and 2 really work for the infinite number of rows in this table?
Is there an infinite number of rows in the first place?
Puzzled,
Karl
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I see. Nobody answered this the way I would have... Taking $u = 2 n+1,$ we are solving $$ u^2 - 8 m^2 = 1. $$ A beginning solution is $(3,1).$ Given a solution $(u,m),$ we get a new one $$ (3 u + 8 m, u + 3 m). $$ Then $n = (u-1)/2$ for each pair.
So, with $n^2 + n = 2 m^2$ and $u = 2 n + 1,$ we get triples
$$ (n,u,m) $$
$$ (1,3,1) $$
$$ (8,17,6) $$
$$ (49,99,35) $$
$$ (288,577,204) $$
$$ (1681,3363,1189) $$
$$ (9800,19601,6930) $$
$$ (57121,114243,40391) $$
$$ (332928,665857,235416) $$
$$ (1940449,3880899,1372105), $$
With my letters, each is a similar sequence, let us use $r$ for "row,"
$$ m_1 = 1, m_2 = 6, \; \; m_{r+2} = 6m_{r+1} - m_r, $$
$$ u_1 = 3, u_2 = 17, \; \; u_{r+2} = 6u_{r+1} - u_r, $$
$$ n_1 = 1, n_2 = 8, \; \; n_{r+2} = 6n_{r+1} - n_r + 2. $$
|
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|
algebra help in AoCP intro to induction In the introduction of mathematical induction, section 1.2.1 of Knuth's Art of Computer Programming, I'm struggling with (4) especially this relation:
$\phi^{n-2} +\phi^{n-1} = \phi^{n-2}(1+\phi)$
Looks like my algebra skills have gone all to rot, if I ever had any, so I am hoping someone can provide a detailed expansion of how these two forms are equal.
|
Distributing and using properties of exponents,
\begin{align*}
\phi^{n - 2} (1 + \phi) &= \phi^{n - 2} \cdot 1 + \phi^{n - 2} \cdot \phi^1 \\
&= \phi^{n - 2} + \phi^{n - 1}
\end{align*}
|
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Inequality from a Sequence I am working on a problem and I am lead to prove the following inequality which is true based on writing out the sequence and on the fact that it should be true based on what I am trying to prove.
Let $s_0,s_1,s_2,....$ be a sequence of positive numbers satisfying $s_0=s_1=1$ and $s_n = s_{n-1} - \alpha s_{n-2}$ where $0<\alpha<\tfrac{1}{4}$. Prove that $2s_{n+1}\geq s_n$.
Perhaps, the following property will help, $s_{n-1}s_{n+1}\leq s_n^2$ that you can assume.
|
Letting $A\lt B$ be the roots of the equation $x^2-x+\alpha=0,$ we have
$$s_{n+1}-As_{n}=B(s_{n}-As_{n-1})=\cdots=B^n(s_1-As_0)=B^n(1-A).$$
$$s_{n+1}-Bs_{n}=A(s_{n}-Bs_{n-1})=\cdots=A^n(s_1-Bs_0)=A^n(1-B).$$
Hence, we have
$$s_n=\frac{B^n(1-A)-A^n(1-B)}{B-A}.$$
Hence,
$$2s_{n+1}-s_n\ge 0$$
$$\iff 2\cdot B^{n+1}(1-A)-2\cdot A^{n+1}(1-B)-B^n(1-A)+A^n(1-B)\ge 0$$
$$\iff B^n(2B-1)(A+1)+A^n(1-2A)(1-B)\ge0$$
This is true because of
$$0\lt A=\frac{1-\sqrt{1-4\alpha}}{2}\lt\frac 12\lt B=\frac{1+\sqrt{1-4\alpha}}{2}\lt 1.$$
Hence, we now know that we prove that $2s_{n+1}-s_n\ge 0$.
|
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|
largest fraction less than 1 What is the mathematically rigorous way to answer the question: "what is the largest fraction less than 1"? (or to explain why it cannot be answered in the manner worded).
|
There is no such fraction.
To see that this is the correct answer, note that for any fraction $\frac{p}{q}$ less than one, there is a slightly bigger fraction which is still less than one; in particular, $\frac{p+q}{2q}$ is a fraction such that $\frac{p}{q} < \frac{p+q}{2q} < 1$.
|
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|
Dimension of space spanned by row vectors Question is to find dimension of spaces spanned by vectors :
$$\alpha_1=(1,1,0,1,0,0),\\
\alpha_2=(1,1,0,0,1,0),\\
\alpha_3=(1,1,0,0,0,1),\\
\alpha_4=(1,0,1,1,0,0),\\
\alpha_5=(1,0,1,0,1,0),\\
\alpha_6=(1,0,1,0,0,1).$$
I tried to make it down to row echelon form but it is not giving clear result... If i am not able to make a row zero that does not mean it can not be done... So, I am unale to conclude anything...
All I can see is that space should be of dimension at least four and it can not be six.
Please give hints to see this in less mechanical way.
Thank you.
|
When we write the given vectors in a matrix form and reduce it to a row echelon thus obtaining the rank as 4 where the 1 St four columns are linearly independent implies that the first four scalars are zero..so out of the 6 vectors given when only 4 are linearly independent..we can say that the vector space is spanned by linearly independent vectors alone that is the number 4.. so dim of the v.sp should be 4.
|
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|
Visualizing a complex valued function of one real parameter I'm looking for a way to capture/graph or visualize it in my head, but I can't find how..
a 2-dimensional path won't do, because it doesn't reveal the rate-of-change..
2 1-dimensional graphs on top of each other doesn't help much either..
3-dimensional space with 2-dimensional slices for values of the parameter is the best I could come up with..
Is there a better way?
|
[> with(plots):
f := z-> exp(I*z):
complexplot3d(f, -2-2*I .. 2+2*I);
Or
[> complexplot(exp(I*x), x = -Pi .. Pi);
|
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Pick's theorem application In a game let us say that we have $3$ ways to score points, getting $1,2 \text{ or } 3$ points. I make a total of $30$ points. What are the various ways I can make $30$ points?
If we plug in values, we have $a + 2b + 3c = 30$ and we have to find ordered pairs $a,b,c$ where $a,b,c$ are whole numbers. However i saw another way of solving this question, using Pick's theorem. I found it better because plugging in would be difficult if it would have been $300$ in place of $30$.
According to the solution I read, the number of $2$s and $3$s will affect the number of $1$s, so we can make represent the $2$s on the x-axis and the $3$s on the y-axis and vice-versa. So how can I use the Pick's theorem to solve this question?
|
I don't know why you would want to use Pick's theorem to solve this, but you are trying to find the number of lattice points satisfying your equation. These lie in a triangle, so if you project it into the $x,y$ plane the number of lattice points stays the same, and you can use use pick's theorem (assuming you can compute the number of boundary points -- this is easier, since dimension is reduced by one.
A completely mechanical way to answer these questions is given by Euler's generating function -- the number of solutions is the coefficient of $x^{30}$ in
$$\frac{1}{1-x} \frac{1}{1-x^2} \frac1{1 - x^3}.$$ You can write down the general term in closed form using partial fractions, but I leave this to you...
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
|
We may as well assume $x\le y\le z$ (and then count rearrangements of the variables as appropriate). The smallest variable, $x$, cannot be greater than $3$ (or else $1/x+1/y+1/z\lt1/3+1/3+1/3=1$), nor can it be equal to $1$ (or else $1/x+1/y+1/z=1+1/y+1/z\gt1$). So either $x=2$ or $x=3$.
If $x=3$, then $y=z=3$ as well (for the same reason as before), which gives the solution $(x,y,z)=(3,3,3)$.
If $x=2$, then $1/2+1/y+1/z=1$ implies
$${1\over2}={1\over y}+{1\over z}$$
Applying the inequality $y\le z$ to this equation, we see that $y$ must be greater than $2$ but cannot be greater than $4$, so $y=3$ or $y=4$. Each of these gives a solution, $(x,y,z)=(2,3,6)$ and $(2,4,4)$.
Counting rearrangements, we get the OP's $10$ solutions and no others.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
}
|
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